Inequalities marathon

MathLinks.ro Inequalities Marathon version 1.00 http://www.mathlinks.ro/viewtopic.php?t=299899 Author: MathLink ers Editor: Hassan Al-Sibyani October 10, 2009 Introduction Hello everyone, this file contain 100 problem in inequalities, generally at Pre-Olympiad level Those problems has been collected from MathLinks.ro at the topic called ”‘Inequalities Marathon”’ I decide to make this work to be a good reference for young students who are interested in inequalities This work provides some of the nicest problems among international Olympiads, as well as some amazing problems by the participants Some of the problems contains more than one solution from the marathon and even outside the marathon from other sources The file description: I- The next page contain all the members that participate with URLs to their MathLinks.ro user-name information and also that page contain some of the real names of the participants that like to have their real names on this file II- Then there are some pages under the title of ”‘Important Definitions”’, this page contain some important definition that you may see them in the problems so if you don’t know one of those definitions you can look at it there III- Then the problems section comes each problem with its solution(s) A- The problem format: ”‘Problem No ‘Author’ (Proposer):”’ B- The solution format: ”‘Solution No (Propser):”’ (Note that the proposer of the solution is not necessary the one who make this solution) I would like to all the participants of this marathon for making this work, also I would like to give a special thank to Endrit Fejzullahu and Popa Alexandru for their contributing making this file If you have another solution to any of those problem or a suggestion or even if you find any mistake of any type on this file please PM me at my MathLinks.ro user or e-mail me at hasan4444@gawab.com Editor Participant sorted alphabetically by thier Mathlink username: Agr 94 Math alex2008 apratimdefermat beautifulliar bokagadha Brut3Forc3 can hang2007 dgreenb801 Dimitris X enndb0x Evariste-Galois FantasyLover geniusbliss great math hasan4444 keyree10 Mateescu Constantin mathinequs Maths Mechanic Mircea Lascu Pain rinnegan peine Potla R.Maths Rofler saif socrates Virgil Nicula Zhero List of names: Name Apratim De Aravind Srinivas Dimitris Charisis Endrit Fejzullahu Hassan Al-Sibyani Hoang Quoc Viet Mateescu Constantin Mharchi Abdelmalek Mircea Lascu Mohamed El-Alami Raghav Grover Toang Huc Khein Popa Alexandru Redwane Khyaoui Sayan Mukherjee Virgil Nicula Vo Quoc Ba Can Mathlink User apratimdefermat Agr 94 Math Dimitris X enndb0x hasan4444 great math Mateescu Constantin Evariste-Galois Mircea Lascu peine Maths Mechanic Pain Rinnegan alex2008 R.Maths Potla Virgil Nicula can hang2007 Important Definitions Abel Summation Formula Let a1 , a2 , · · · , an and b1 , b2 , · · · , bn be two finite sequances of number Then a1 b1 +a2 b2 + +an bn = (a1 −a2 )b1 +(a2 −a3 )(b1 +b2 )+· · ·+(an−1 −an )(b1 +b2 +· · ·+bn−1 )+an (b1 +b2 +· · ·+bn ) AM-GM (Arithmetic Mean-Geometric Mean) Inequality If a1 , a2 , · · · , an are nonnegative real numbers, then √ a1 , a2 + · · · + an ≥ n a1 a2 · · · an n with equality if and only if a1 = a2 = · · · = an Cauchy-Schwarz Inequality For any real numbers a1 , a2 , · · · , an and b1 , b2 , · · · , bn (a21 + a22 + · · · + a2n )(b21 + b22 + · · · + b2n ) ≥ (a1 b1 + a2 b2 + · · · + an bn )2 with equality if and only if and bi are proportional, i = 1, 2, · · · , n Chebyshev’s Inequality If (a1 , a2 , · · · , an ) and (b1 , b2 , · · · , bn ) are two sequences of real numbers arranged in the same order then : n(a1 b1 + a2 b2 + + an bn ) ≥ (a1 + a2 + + an )(b1 + b2 + + bn ) Cyclic Sum Let n be a positive integer Given a function f of n variables, define the cyclic sum of variables (a1 , a2 , · · · , an ) as f (a1 , a2 , a3 , an ) = f (a1 , a2 , a3 , an )+f (a2 , a3 , a4 , an , a1 )+f (a3 , a4 , an , a1 , a2 )+ .+f (an , a1 , a2 , an−1 ) cyc QM-AM-GM-HM (General Mean) Inequality If a1 , a2 , · · · , an are nonnegative real numbers, then √ a21 + a22 + · · · + a2n a1 , a2 + · · · + an ≥ ≥ n a1 a2 · · · an ≥ n n with equality if and only if a1 = a2 = · · · = an a1 + a2 n + ··· + an GM-HM (Geometric Mean-Harmonic Mean) Inequality If a1 , a2 , · · · , an are nonnegative real numbers, then √ n a1 a2 · · · an ≥ a1 + a2 n + ··· + an with equality if and only if a1 = a2 = · · · = an Hlawka’s Inequality Let z1 , z2 , z3 be three complex numbers Then: |z1 + z2 | + |z2 + z3 | + |z3 + z1 | ≤ |z1 | + |z2 | + |z3 | + |z1 + z2 + z3 | Holder’s Inequality Given mn (m, n ∈ N) positive real numbers (xij )(i1, m , j1, n) , then : m  n m ≥ xij   i1  m1 m j1 j1 Jensen’s Inequality Let F be a convex function of one real variable a1 + · · · + an = Then: xijm i1 Let x1 , , xn ∈ R and let a1 , , an ≥ satisfy F (a1 x1 + · · · + an xn ) ≤ a1 F (x1 ) + · · · + an F (xn ) Karamata’s Inequality Given that F (x) is a convex function in x, and that the sequence {x1 , , xn } majorizes the sequence {y1 , , yn }, the theorem states that F (x1 ) + · · · + F (xn ) ≥ F (y1 ) + · · · + F (yn ) The inequality is reversed if F (x) is concave, since in this case the function −F (x) is convex Lagrange’s Identity The identity: n ak bk k=1 n n a2k = k=1 b2k k=1 − (ak bj − aj bk )2 1≤k≤j≤n applies to any two sets {a1 , a2 , · · · , an } and {b1 , b2 , · · · , bn } of real or complex numbers Minkowski’s Inequality For any real number r ≥ and any positive real numbers a1 , a2 , · · · , an and b1 , b2 , · · · , bn Then 1 ((a1 + b1 )r + (a2 + b2 )r + · · · + (an + bn )r ) r ≤ (ar1 + ar2 + · · · + arn ) r + (br1 + br2 + · · · + brn ) r Muirhead’s Inequality If a sequence A majorizes a sequence B, then given a set of positive reals x1 , x2 , · · · , xn x1 a1 x2 a2 · · · xn an ≥ sym x1 b1 x2 b2 · · · xn bn sym QM-AM (Quadratic Mean-Arithmetic Mean) Inequality Also called root-mean-square and state that if a1 , a2 , · · · , an are nonnegative real numbers, then a21 + a22 + · · · + a2n a1 , a2 + · · · + an ≥ n n with equality if and only if a1 = a2 = · · · = an The Rearrangement Inequality Let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤ · · · ≤ bn be real numbers For any permutation (a1 , a2 , · · · , an ) of (a1 , a2 , · · · , an ), we have a1 b1 + a2 b2 + · · · + an bn ≤ a1 b1 + a2 b2 + · · · + an bn ≤ an b1 + an−1 b2 + · · · + a1 bn with equality if and only if (a1 , a2 , · · · , an ) is equal to (a1 , a2 , · · · , an ) or (an , an−1 , · · · , a1 ) respectively Schur’s Inequality For all non-negative a, b, c ∈ R and r > 0: ar (a − b)(a − c) + br (b − a)(b − c) + cr (c − a)(c − b) ≥ The four equality cases occur when a = b = c or when two of a, b, c are equal and the third is Common Cases The r = case yields the well-known inequality:a3 + b3 + c3 + 3abc ≥ a2 b + a2 c + b2 a + b2 c + c2 a + c2 b When r = 2, an equivalent form is: a4 + b4 + c4 + abc(a + b + c) ≥ a3 b + a3 c + b3 a + b3 c + c3 a + c3 b Weighted AM-GM Inequality For any nonnegative real numbers a1 , a2 , · · · , an if w1 , w2 , · · · , wn are nonnegative real numbers (weights) with sum 1, then wn w2 w1 a1 + w2 a2 + · · · + wn an ≥ aw a2 · · · an with equality if and only if a1 = a2 = · · · = an Problem ‘India 2002’ (Hassan Al-Sibyani): For any positive real numbers a, b, c show that the following inequality holds a b c c+a a+b b+c + + ≥ + + b c a c+b a+c b+a First Solution (Popa Alexandru): Ok After not so many computations i got that: a b c a+b b+c c+a + + − − − b c a c+a a+b b+c = abc (a + b)(b + c)(c + a) + b2 c2 a b a2 c + + − − − 2 b c a b c a abc (a + b)(b + c)(c + a) bc ca ab + + −3 c2 a b a2 b2 c2 a b c ab bc ca So in order to prove the above inequality we need to prove + + ≥ + + and + + ≥ b c a b c a c a b The second inequality is obvious by AM-GM , and the for the first we have: a2 b2 c2 + + b2 c2 a2 ≥3 a2 b2 c2 + + b2 c2 a2 ≥ c a b + + b c a a b c where i used AM-GM and the inequality 3(x2 + y + z ) ≥ (x + y + z)2 for x = , y = , z = b c a So the inequality is proved a b c Second Solution (Raghav Grover): Substitute = x, = y, = z So xyz = The inequality after b c a substitution becomes x2 z + y x + z x + x2 + y + z ≥ x + y + z + x2 z + y x + z x ≥ So now it is left to prove that x2 + y + z ≥ x + y + z which is easy Third Solution (Popa Alexandru): Bashing out it gives a4 c2 + b4 a2 + c4 b2 + a3 b3 + b3 c3 + a3 c3 ≥ abc(ab2 + bc2 + ca2 + 3abc) which is true because AM-GM gives : a3 b3 + b3 c3 + a3 c3 ≥ 3a2 b2 c2 and by Muirhead : a4 c2 + b4 a2 + c4 b2 ≥ abc(ab2 + +bc2 + ca2 ) Fourth Solution (Popa Alexandru): Observe that the inequality is equivalent with: cyc a2 + bc ≥3 a(a + b) Now use AM-GM: cyc (a2 + bc) a2 + bc ≥33 a(a + b) abc (a + b) So it remains to prove: (a2 + bc) ≥ abc (a + b) Now we prove (a2 + bc)(b2 + ca) ≥ ab(c + a)(b + c) ⇔ a3 + b3 ≥ ab2 + a2 b ⇔ (a + b)(a − b)2 ≥ Multiplying the similars we are done Problem ‘Maxim Bogdan’ (Popa Alexandru): Let a, b, c, d > such that a ≤ b ≤ c ≤ d and abcd = Then show that: (a + 1)(d + 1) ≥ + 4d Solution (Mateescu Constantin):From the condition a ≤ b ≤ c ≤ d we get that a ≥ + (d + 1) d3 Now let’s prove that + (d + 1) ≥ + d 4d This is equivalent with: (d3 + 1)(d + 1) ≥ 3d3 + 2 ⇐⇒ [d(d − 1)] − [d(d − 1)] + ≥ ⇐⇒ d(d − 1) − 21 ≥ √ 1+ 1 Equality holds for a = and d(d − 1) − = ⇐⇒ d = d 2 d3 =⇒ (a + 1)(d + 1) ≥ Problem ‘Darij Grinberg’ (Hassan Al-Sibyani): If a, b, c are three positive real numbers, then a (b + c) + b (c + a) + c (a + b) ≥ (a + b + c) First Solution (Dimitris Charisis): a2 (a + b + c)2 ≥ 2 ab + ac + 2abc sym a b + 6abc So we only have to prove that: 4(a + b + c)3 ≥ a2 b + 54abc ⇐⇒ 4(a3 + b3 + c3 ) + 12 sym 3 sym a b ≥ 30abc 4(a + b + c ) + sym a2 b ≥ 6abc and a3 + b3 + c3 ≥ 3abc But a2 b + 24abc ≥ sym So 4(a3 + b3 + c3 ) + a2 b ≥ 30abc sym a2 b + 54abc ⇐⇒ sym Second Solution (Popa Alexandru): Use Cauchy-Schwartz and Nesbitt: (a + b + c) b c a + + (b + c)2 (c + a)2 (a + b)2 a b c + + b+c c+a a+b ≥ ≥ Problem ‘United Kingdom 1999’ (Dimitris Charisis): For a, b, c ≥ and a + b + c = prove that 7(ab + bc + ca) ≤ + 9abc First Solution (Popa Alexandru): Homogenize to 2(a + b + c)3 + 9abc ≥ 7(ab + bc + ca)(a + b + c) Expanding it becomes : a3 + sym a2 b + 21abc ≥ sym a2 b + 21abc sym So we just need to show: a3 ≥ sym a2 b sym which is obvious by a3 + a3 + b3 ≥ 3a2 b and similars Second Solution (Popa Alexandru): Schur gives + 9abc ≥ 4(ab + bc + ca) and use also 3(ab + bc + ca) ≤ (a + b + c)2 = Suming is done Problem ‘Gheorghe Szollosy, Gazeta Matematica’ (Popa Alexandru): Let x, y, z ∈ R+ Prove that: x(y + 1) + y(z + 1) + z(x + 1) ≤ (x + 1)(y + 1)(z + 1) Solution (Endrit Fejzullahu): Dividing with the square root on the RHS we have : x + (x + 1)(z + 1) y + (x + 1)(y + 1) z ≤ (y + 1)(z + 1) By AM-GM x ≤ (x + 1)(z + 1) x + x+1 y+1 y ≤ (x + 1)(y + 1) y + y+1 x+1 z ≤ (y + 1)(z + 1) z + z+1 y+1 Summing we obtain LHS ≤ x + x+1 x+1 + y + y+1 y+1 + z + z+1 z+1 = Problem ‘Romanian Regional Mathematic Olympiad 2006’ (Endrit Fejzullahu): Let a, b, c be positive numbers , then prove that 1 4b 4c 4a + + + + ≥ a b c 2a + b2 + c2 a + 2b2 + c2 a + b2 + 2c2 √ First Solution (Mateescu Constantin): By AM − GM we have 2a2 + b2 + c2 ≥ 4a bc 4a 4a ≤ √ =√ =⇒ 2a2 + b2 + c2 4a bc bc 1 Addind the similar inequalities =⇒ RHS ≤ √ + √ + √ (1) ca ab bc 2 1 1 1 Using Cauchy-Schwarz we have √ + √ + √ ≤ + + a b c ca ab bc 1 1 1 So √ + √ + √ ≤ + + (2) a b c ca ab bc From (1), (2) we obtain the desired result Second Solution (Popa Alexandru): By Cauchy-Schwatz : 2a2 Then we have RHS ≤ cyc 4a a a ≤ + 2 2 +b +c a +b a + c2 a+b ≤ a2 + b2 cyc ≤ a+b cyc 1 + 2a 2b = LHS Third Solution (Popa Alexandru): Since 2a2 + b2 + c2 ≥ a2 + ab + bc + ca = (a + b)(c + a) There we have: 8(ab + bc + ca) 1 4a = ≤ + + LHS ≤ (a + b)(c + a) (a + b)(b + c)(c + a) a b c cyc The last one is equivalent with (a + b)(b + c)(c + a) ≥ 8abc Problem ‘Pham Kim Hung’ (Mateescu Constantin): Let a, b, c, d, e be non-negative real numbers such that a + b + c + d + e = Prove that: abc + bcd + cde + dea + eab ≤ Solution (Popa Alexandru): Assume e ≤ min{a, b, c, d} Then AM-GM gives : e(c + a)(b + d) + bc(a + d − e) ≤ e(5 − e)2 (5 − 2e)2 + ≤5 27 the last one being equivalent with: (e − 1)2 (e + 8) ≥ And cyc (xy + yz + zx)2 (xy + yz + zx)2 yz ≥ = ≥ ⇐⇒ (xy + yz + zx)2 ≥ 3xyz 2 +x x yz + y xz + z xy + 3xyz 4xyz x2 This is true since x + y + z = and (xy + yz + zx)2 ≥ 3xyz(x + y + z) Second Solution (Endrit Fejzullahu): Let us observe that by Cauchy-Schwartz we have : cyc x + yz = x2 + x cyc x(x + y + z) + yz = x(x + 1) cyc (x + y)(x + z) = x(x + y + x + z) x cyc x+y + x x+z ≥ (1 + + 1)2 = =3 3 Then we can conclude that : LHS = cyc x2 + x − x2 + x x + yz ≤3−3=0 x2 + x cyc Problem 70 ‘Claudiu Mandrila’ (Endrit Fejzullahu): Let a, b, c > such that abc = Prove that : b10 c10 a7 b7 c7 a10 + + ≥ + + 7 b+c c+a a+b b +c c +a a + b7 Solution (Sayan Mukherjee): Since abc = so, we have: a10 = b+c a7 b3 c3 (b + c) ≥ 64 (b a7 + c)6 · (b + c) So we are only required to prove that: (b + c)7 ≤ 26 b7 + 26 c7 But, from Holder; (b7 + c7 ) (1 + 1) ≥ b + c Hence we are done Problem 71 ‘Hojoo Lee, Crux Mathematicorum’ (Sayan Mukherjee): Let a, b, c ∈ R+ ;Prove that: 9(a + b + c)2 (a + b3 + c3 ) + ≥ 33 abc a + b2 + c2 First Solution (Endrit Fejzullahu): Inequality is equivalent with : (a + b + c)(a2 + b2 + c2 ) − 9abc (a − b)2 + (b − c)2 + (c − a)2 ≥ Second Solution (Popa Alexandru): LHS − RHS = a2 b2 c2 + + −3 bc ca ab 2− 40 18abc (a + b + c)(a2 + b2 + c2 ) ≥0 Problem 72 ‘Cezar Lupu’ (Endrit Fejzullahu): Let a, b, c be positive real numbers such that a + b + c ≥ a b c b + c + a Show that : a3 c b3 a c3 b + + ≥ b(c + a) c(a + b) a(b + c) First Solution (Sayan Mukherjee): Observe that a + b + c ≥ from rearrangement inequality: a; b; c 1 b; c; a a; b; c ≥ + b c + c a ≥ (1) from AM-GM Also a b c a b c + + ≥ + + b c a c a b =⇒ 1 c; a; b a b (2) Hence we rewrite LHS of the inequality in the form: cyc a2 b(a + c) · ac = cyc a2 b a + b c ≥ (a + b + c)2 b cyc c + cyc c b ≥ (a + b + c)2 (a + b + c) ≥ ≥ a cyc b 2 (From (1) (2)) QED Second Solution (Popa Alexandru): Holder gives : cyc a3 c 2(a + b + c) b(c + a) Therefore we have : cyc a b c + + b c a ≥ (a + b + c)3 a3 c (a + b + c)3 ≥ b(c + a) 2(a + b + c) ab + cb + c a ≥ Third Solution (Popa Alexandru): We have that : cyc a3 c = b(c + a) cyc b a b 1 c + a By Cauchy-Schwartz we get : cyc b 1 + c a cyc b a b 1 c + a ≥ So: cyc b a b 1 c + a a cyc b ≥ It’ll be enough to prove that : cyc a ≥ b 41 cyc ab a cyc b cyc ab cyc a b By contradiction method we assume that cyc a≥2 this with the condition cyc cyc cyc But using AM-GM as x2 + cyc which is equivalent with ab a , we will have that b a > a b2 + a < b b + b c2 + c + c a2 + ab2 < cyc a Suming cyc a 1 1 = x2 + + ≥ x2 · · = 3, we get 3(a + b + c) > 3(a + b + c), x x x x x contradiction Problem 73 ‘Popa Alexandru’ (Sayan Mukherjee): For x, y, z > and x + y + z = Find Pmin if : P = y3 z3 x3 + + 2 (1 − x) (1 − y) (1 − z)2 Solution (Zhero): (3x − 1)2 ≥ ⇐⇒ x3 ≥ x − Adding these up yields that the RHS is ≥ − = (1 − x) 4 Problem 74 ‘Hlawka’ (Zhero): Let x, y, and z be vectors in Rn Show that |x + y + z| + |x| + |y| + |z| ≥ |x + y| + |y + z| + |z + x| First Solution (Dimitris Charisis): I will solve this inequality for x, y, z ∈ C First observe that |x + y + z|2 + |x|2 + |y|2 + |z|2 = |x + y|2 + |y + z|2 + |z + x|2 To prove this just use the well-known identity |z|2 = z · z¯ Then our inequality is: |x + y + z|2 + |x|2 + |y|2 + |z|2 + 2(|x + y + z|)(|x| + |y| + |z|) + |xy| + |yz| + |xz| ≥ |x + y|2 + From we only need to prove that: (|x + y + z|)(|x| + |y| + |z|) + |xy| + |yz| + |xz| ≥ |x + y||y + z| But |(x + y)(y + z)| = |y(x + y + z) + xz| ≤ |y(x + y + z)| + |yz| Cyclic we have the result |x + y y + z| Second Solution (Popa Alexandru): Use Hlawka’s identity : (|a| + |b| + |c| − |b + c| − |c + a| − |a + b| + |a + b + c|) · (|a| + |b| + |c| + |a + b + c|) = = (|b|+|c|−|b+c|)·(|a|−|b+c|+|a+b+c|)+(|c|+|a|−|c+a|)·(|b|−|c+a|+|a+b+c|)+(|a|+|b|−|a+b|)·(|c|−|a+b|+|a+b+c|) ≥ and the conclusion follows 42 Problem 75 ‘Greek Mathematical Olympiad’ (Dimitris Charisis): If x, y, z ∈ R so that x2 + 2y + z = 2 a , a > 0,prove that |x − y + z| ≤ a Solution (Popa Alexandru): The inequality is equivalent with : a2 ≥ x2 + y + z − 2yz − 2xy + 2zx Since the condition this rewrites to : 5 x + 5y + z ≥ x2 + y + z − 2yz − 2xy + 2zx 2 which is equivalent with : (x − y)2 + √ x √ +y 2 + √ z √ +y 2 ≥0 Problem 76 ‘Vasile Cirtoaje and Mircea Lascu, Junior TST 2003, Romania’ (Hassan Al-Sibyani): Let a, b, c be positive real numbers so that abc = Prove that: 1+ ≥ a+b+c ab + ac + bc First Solution (Endrit Fejzullahu): Given inequality is equivalent with : ab + bc + ca ≥ 6(a + b + c) 36(a + b + c)2 =⇒ (ab + bc + ca)2 ≥ a+b+c+3 (a + b + c + 3)2 By the AM-GM inequality we know that : (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c) It is enough to prove that 3(a + b + c) ≥ 36(a + b + c)2 ⇐⇒ (a + b + c − 3)2 ≥ (a + b + c + 3)2 Second Solution (Dimitris Charisis): Setting a = 1 , b = , c = we have to prove that: x y z ≥ xy + yz + zx x+y+z But ≥ xy + yz + zx (x + y + z)2 So we have to prove that: 1+ ≥ (x + y + z)2 x+y+z Now setting x + y + z = k, the inequality becomes: 1+ 43 (k − 3)2 ≥ which is true Third Solution (Popa Alexandru): We have (ab + bc + ca)2 ≥ 3abc(a + b + c) = 3(a + b + c) Then 1+ ≥ ≥1+ a+b+c (ab + ac + bc)2 ab + ac + bc Problem 77 ‘Vasile Pop, Romania 2007 Shortlist’ (Endrit Fejzullahu): Let a, b, c ∈ [0, 1] Show that : a b c + + + abc ≤ + bc + ca + ab Solution (Dimitris Charisis): bc ≤ abc a a So ≤ + bc + abc a+b+c a cyclic we have that ≤ + bc + abc So our problem reduces to: a+b+c + abc ≤ But a + b + c ≤ abc + + abc Because: (1 − ab)(1 − c) ≥ ⇐⇒ − c − ab + abc ≥ ⇐⇒ + abc ≥ ab + c + So we have to prove that ab + c + ≥ a + b + c But (a − 1)(b − 1) ≥ ⇐⇒ ab − a − b + ≥ ⇐⇒ ab + ≥ a + b ⇐⇒ ab + c + ≥ a + b + c a+b+c abc + So ≤ =1+ + abc abc + abc + Finally our inequality reduces to: + abc ≤ abc + Setting abc = x ≤ we have: + x ≤ ⇐⇒ 2x2 − x − ≤ ⇐⇒ (x − 1)(2x + 1) ≤ which is true x+1 Problem 78 ‘Greek 2004’ (Dimitris Charisis): Find the best constant M so that the inequality: x4 + y + z + xyz(x + y + z) ≥ M (xy + yz + zx)2 for all x, y, z ∈ R Solution (Endrit Fejzullahu): Setting x = y = z = ,we see that M ≤ 2 We prove that the inequality is true for M = 3 x4 + y + z + xyz(x + y + z) ≥ 44 (xy + yz + zx)2 After squaring (xy + yz + zx)2 on the RHS ,Inequality is equivalent with : x4 + y + z + x4 + y + z − (xy)2 − (yz)2 − (zx)2 ≥ xyz(x + y + z) It is obvious that x4 + y + z ≥ (xy)2 + (yz)2 + (zx)2 ,so it is enough to prove that : x4 + y + z ≥ xyz(x + y + z) x4 + y + z ≥ (xy)2 + (yz)2 + (zx)2 ≥ x2 yz + xy z + xyz = xyz(x + y + z) The last one is true ,setting xy = a, yz = b, zx = c ,it becomes a2 + b2 + c2 ≥ ab + bc + ca ,which is trivial Problem 79 ‘Dragoi Marius and Bogdan Posa’ (Endrit Fejzullahu): Let a, b, c > auch that abc = 1, Prove that : a2 (b5 + c5 ) + b2 (a5 + c5 ) + c2 (b5 + a5 ) ≥ 2(a + b + c) Solution (Sayan Mukherjee): Lemma: Note that a5 + a5 + a5 + b5 + b5 ≥ 3a3 b2 a5 + a5 + b5 + b5 + b5 ≥ 3a2 b3 Summing up; a5 + b5 ≥ a2 b2 (a + b) So LHS = a2 (b5 + c5 ) ≥ a2 b2 c2 (c + b) = 2(a + b + c) from abc = and our lemma above Problem 80 ‘Russia 1995’ (Sayan Mukherjee): For positive x, y prove that: x y ≥ + 2 xy x +y x + y4 Solution (FantasyLover): By AM-GM, we have x4 + y ≥ 2x2 y and x2 + y ≥ 2xy x y x y Hence, + ≤ + = , and we are done x +y x +y 2x y 2xy xy Problem 81 ‘Adapted after an IMO problem’ (FantasyLover): Let a, b, c be the side lengths of a triangle with semiperimeter of Prove that < ab + bc + ca − abc ≤ 28 27 Solution (Sayan Mukherjee): Since a, b, c → sides of a triangle; write: a = x + y; b = y + z; c = z + x so that x + y + z = The inequality is equivalent to : 45 1< (1 − x)(1 − y) − (x + y) ≤ 28 =⇒ < − 27 left side is proved And, since x + y + z = and xyz ≤ x + xyz = + xyz ≤ x+y+z Problem 82 ‘Latvia 2002’ (Sayan Mukherjee): Let a, b, c, d > 0; Prove that: = 28 Since xyz > So the 27 hence the right side is proved 27 1 1 + + + =1 + a4 + b4 + c4 + a4 abcd ≥ First Solution (Zhero): 1 1 Let w = ,x = ,y = , and z = 4 1+a 1+b 1+c + d4 4 Then a = − 1, b = − 1, c = − 1, and d = − w x y z Let f (n) = − ln( − 1) It is easy to verify that f has exactly one inflection point Since we see seek to n minimize f (w) + f (x) + f (y) + f (z), by the inflection point theorem, it suffices to minimize it in the case in which w = x = y In other words, in our original inequality, it suffices to minimize this in the case that a = b = c = But we Let p = a4 , q = b4 , r = c4 , and s = d4 Then we want to show that pqrs ≥ 81 when + p cyc only need to check this in the case in which p = q = r, that is, when s = In other words, we want to p−2 3p3 ≥ 81 ⇐⇒ 3p3 ≥ 81p − 162 ⇐⇒ 3(x − 3)2 (x + 6) ≥ 0, as desired show that p−2 Second Solution (Sayan Mukherjee): d4 1 =1− = ≥ 4 1+d 1+d + a4 (AM-GM) + a4 ) Similarly we get three ther relations and multiplying we get: a,b,c a,b,c (1 a4 b4 c4 d4 ≥ a,b,c,d (1 + a ) 34 =⇒ abcd ≥ a,b,c,d (1 + a ) Problem 83 ‘Vasile Cartoaje and Mircea Lascu’ (Zhero): Let a, b, c, x, y, and z be positive real numbers such that a + x ≥ b + y ≥ c + zanda + b + c = x + y + z Show that ay + bx ≥ ac + xz Solution (mathinequs): a(y − c) + x(b − z) = a(a − x) + (a + x)(b − z) Now i used : (a − x)2 + a2 − x2 a(a − x) = 46 So i got that : (a − x)2 + (a + x)(a − x + 2(b − z)) So i have to prove that : (a + x)(a − x + 2(b − z)) ≥ but a + x is greater or equal than and a − x + 2(b − z) = b + y − c − z which is greater or equal than a(a − x) + (a + x)(b − z) = Problem 84 ‘Nicolae Paun’ (mathinequs): Let a, b, c, x, y, z reals with a + b + c = x2 + y + z = To prove: a(x + b) + b(y + c) + c(z + a) < Solution (Mharchi Abdelmalek): Using AM-GM inequality we have: a2 +x2 +b2 +y +c2 +z ≥ 2ax+2by+2cz Hence we have : (a + b + c)2 + x2 + y + z 1= = (a2 + x2 ) + ab + bc + ca ≥ ax + by + cz + ab + bc + ca = 2 cyc = a(x + b) + b(y + c) + c(z + a) Problem 85 ‘Asian Pacific Mathematics Olympiad’ (Mharchi Abdelmalek): Let x, y, z, be a positive real numbers Prove that: y z 2(x + y + z) x +1 +1 +1 ≥2+ √ xyz y z x Solution (mathinequs): Bashing the left hand side we need to prove : 2(x + y + z) y x + ≥ √ xyz y x But this is AM-GM as : x y 3x + ≥ √ xyz y z Done Problem 86 ‘mateforum.ro’ (mathinequs): Let x, y, z > such that xy + yz + zx = Show: √ 1 + + ≥ x + x3 y + y3 z + z3 Solution (Apartim De): The condition xy + yz + zx = accomodates the substitution x = cot A, y = cot B, z = cot C, where A, B, C are the angles of ∆ABC We obtain the equivalent inequality √ ⊗ cot A(1 + cot2 A) cyc 47 √ ⇔ tan A sin A cyc Let f (x) = tan x sin2 x • f (x) = sin2 x + tan2 x > 0, ∀x ⇔ f (x) is increasing  > 0, ∀x ∈ (0, π/2)  < 0, ∀x ∈ (π/2, π) f (x) = sin x cos x + cos3 x   > 0, ∀x ∈ (π, 3π/2) If the ∆ABC be acute ,√by Jensen, π = f (A) 3f cyc If the ∆ABC be obtuse-angled(at A) , π π f (A) + f (B) + f (C) = f A + + f (B − k) + f C − −k 2 π π A + + (B − k) + C − − k π = 3f 3f 3 f or some suitablek min{B, C} Problem 87 ‘—’ (Apartim De): a, b, c are sides of a triangle.Prove that √ 2b2 a b c +√ +√ 2 2 2 + 2c − a 2c + 2a − b 2a + 2b2 − c2 √ First Solution (Endrit Fejzullahu): According to the Holder’s inequality we have : cyc a √ 2b + 2c2 − a2 a(2b2 + 2c2 − a2 ) ≥ (a + b + c)3 cyc We only need to prove that (a + b + c)3 ≥ a(2b2 + 2c2 − a2 ) cyc It can be rewritten : (abc − (a + c − b)(b + c − a)(a + b − c)) + a3 + b3 + c3 − 3abc ≥ By Schur abc ≥ (a + c − b)(b + c − a)(a + b − c) and by AM-GM a3 + b3 + c3 ≥ 3abc ,so we’re done ! Second Solution (geniusbliss): √ √ √ a 3a(2b2 + 2c2 − a2 ) 3a a √ √ + + ≥3 =3 3 2 2 2 (a + b + c) a+b+c 2b + 2c − a 2b + 2c − a cyclic cyclic cyclic cyclic and finish off the problem 48 Problem 88 ‘Pavel Novotn, Slovakia’ (Endrit Fejzullahu): Let a,b,c,d be positive real numbers such that a b c d abcd = and a + b + c + d > + + + Prove that b c d a a+b+c+d< c d a b + + + a b c d First Solution (mathinequs): It’s AM-GM ! ad + ab + cb ≥ 4a Suming the other and with condition : 3(a + b + c + d) + ab + cb + dc + ad > ab + cb + dc + ad + ab + cb + dc + ad ≥ 4(a + b + c + d) Second Solution (socrates): The second condition implies that a+b+c+d> a b c d a2 b2 c2 d2 (a + b + c + d)2 + + + ≥ + + + ≥ b c d a ab bc cd da ab + bc + cd + da so ab + bc + cd + da > a + b + c + d Now, it is b c d a (bc)2 (cd)2 (da)2 (ab)2 (bc)2 (da)2 (cd)2 (ab)2 + + + = + + + = c + d + a + b > 2 2 a b c d abc bcd cda dab d b a c (bc + cd + da + ab)2 (a + b + c + d)2 > =a+b+c+d a b c d a+b+c+d b + c + d + a completing the proof Problem 89 ‘—’ (geniusbliss): Let a1 , a2 , a3 , an and b1 , b2 , b3 , , bn be real numbers such that - a1 ≥ a1 + a2 a1 + a2 + a3 a1 + a2 + a3 + + an b1 + b2 b1 + b2 + b3 b1 + b2 + b3 + + bn ≥ ≥ ≥ , b1 ≥ ≥ ≥ ≥ n n , then prove that a1 b1 + a2 b2 + a3 b3 + + an bn ≥ · (a1 + a2 + a3 + + an ) · (b1 + b2 + b3 + + bn ) n Solution (geniusbliss): For each k1, 2, , n,to we denote Sk = a1 + a2 + a3 + + ak and bn+1 = Then by Abel’s Formulae, we have n n n Si bi = (bi − bi+1 )Si = i(bi − bi+1 ) i i=1 i=1 i=1 According to Abel’s Formulae, n n−1 n n S2 S3 Sn−1 Sn S2 bi = (S1 − )(b1 −b2 )+( − )(b1 +b2 −2b3 )+ +( − )( −(n−1)bn )+ ( a1 )( b1 ) 2 n − n i=1 n i=1 i=1 i=1 S1 S2 Sn By Hypothesis, we have ≥ ≥ ≥ so its enough to prove that n k bi ≥ kbk+1 i=1 This one comes directly from the hypothesis k k+1 bi ≥ bi k i=1 i=1 49 Problem 90 ‘mateforum.ro’ (mathinequs): Let a, b, c positives with abc(a + b + c − abc) ≤ = Show : + a2 Solution (socrates): Multiplying out we get 2a2 b2 c2 + a2 b2 + b2 c2 + c2 a2 = Moreover, abc(a+b+c−abc) = 1 2abc(a + b + c) − 2a2 b2 c2 = 2abc(a + b + c) + a2 b2 + b2 c2 + c2 a2 − = (ab + bc + ca)2 − 2 so it is enough to prove that ab + bc + ca ≤ x y z Now put a2 = , b2 = , c2 = , x, y, z > Then y+z x+z x+y xy x y ≤ + and similarly ab = (y + z)(x + z) x+z y+z z y yz ≤ + bc = (x + z)(x + y) x+z x+y z xz x ac = ≤ + (y + z)(x + y) x+y y+z Adding these, we are done Problem 91 ‘Komal’ (Hassan Al-Sibyani): For arbitrary real numbers a, b, c Prove that: √ a2 + (1 − b)2 + b2 + (1 − c)2 + c2 + (1 − a)2 ≥ First Solution (Dimitris Charisis): By Minkowski’s inequality we have: LHS ≥ (a + b + c)2 + (a + b + c − 3)2 Setting a + b + c = x our inequality becomes: √ 9 2 x + (x − 3) ≥ ⇐⇒ 2x2 − 6x + ≥ ⇐⇒ x2 − 3x + ≥ ⇐⇒ 2 x− 2 ≥0 ,which is true Second Solution (Mharchi Abdelmalek): we have the flowing well-know inequality for all real numbers a, b : |a + − b| √ a2 + (1 − b)2 ≥ Simarly for the other numbers.Hence we have √ |a − b + 1| + |b − c + 1| + |c − a + 1| |a − b + b − c + c − a + 3| √ √ LHS ≥ ≥ = 2 50 Problem 92 ‘Asian Pacific Mathematics Olympiad 1996’ (Dimitris Charisis): For a, b, c sides of a triangle prove that: √ √ √ √ √ √ a + b − c + a − b + c + −a + b + c ≤ a + b + c First Solution (geniusbliss): substitute a = x√+ y, b = y + z, c = z + x we have to prove then, √ √ √ √ x + y + y + z + z + x ≥ 2x + 2y + 2z which is true because squaring both sides, √ √ √ (x + y)(y + z) ≥ xy + yz + zx which is true by Cauchy - (x + y)(y + z) ≥ we have to prove cyc √ √ ( xy + yz)2 and similarly for others ,equality holds for x = y = z x+y z+x and b = y+z Second Solution (Redwane Khyaoui): First of all , we set a = and c = √ √ √ √ √ √2 √ the inequality become into : 2( x + y + z) ≤ x + y + y + z + z + x √ √ √ √ then we use the well-known inequality : ( x + y)2 ≤ x + y which means 2(x + y) ≥ x + y and then summing the three inequalities , gives inequality desired Problem 93 ‘geniusbliss’ (geniusbliss): Prove that for x, y, z positive reals such as xz ≥ y , xy ≥ z the follwing inequality holds27 (x2 − yz)2 ≥ (xz − y )(xy − z ) Solution (geniusbliss): this obviously implies the LHS is also positive y z x make this substitution - a = , b = , c = , y z x the inequality gets transformed into 8(a − c)2 ≥ 27(a − b)(b − c) or 2((a − c)2 ) 13 ≥ 3((a − b)(b − c)) we have a ≥ b ≥ c, multiply both sides by (a − c) , and we get, 2(a−c) ≥ 3((a−b)(b−c)(a−c)) but this is just AM-GM if we write the LHS as 2(a−c) = a−b+b−c+a−c so we are done and the inequality is true with equality holding when a = b = c Problem 94 ‘Kvant 1988’ (Mharchi Abdelmalek): Let a, b, c ≥ such that a4 +b4 +c4 ≤ 2(a2 b2 +b2 c2 +c2 a2 ) Prove that : a2 + b2 + c2 ≤ 2(ab + bc + ca) Solution (Hassan Al-Sibyani): The condition a4 ≤ 51 a2 b2 is equivalent to (a + b + c)(a + b − c)(b + c − a) ≥ In any of the cases a = b + c, b = c + a, c = a + b, the inequality a2 ≤ ab is clear So, suppose a = b+c, b = c+a, c = a+b Because at most one of the numbers b+c−a, c+a−b, a+b−c is negative and their products is non-negative, all of them are positive Thus, we may assume that: a2 < ab + ac, b2 < bc + ba, c2 < ca + cb and the conclusion follows Problem 95 ‘Junior Balkan Mathematical Olympiad 2002 Shortlist’ (Hassan Al-Sibyani): Let a, b, c be positive real numbers Prove that b3 c3 a2 b2 c2 a3 + 2+ ≥ + + b c a b c a First Solution (Redwane Khyaoui): If a ≥ b ≥ c then we put a = tb and b = t c so a = tt c with t, t ≥ th inequality become into t5 t + t t2 + ≥ t4 t + t t2 + tt and we consider the fonction f (t) = t5 t + t t2 + − t4 t + t t2 + tt so f (t) = 5t t4 − 4t t3 + 2t(t − t ) − tt + then , since we know that t, t ≥ so f (t) ≥ which means f is increasing function , t ≥ → f (t) ≥ f (1) = t − t − t + setting again g(t ) = t − t − t + so g (t) = 5t − 4t − ≥ because t ≥ whic means that g is increasing fonction , so t ≥ → g(t) ≥ g(1) = and finally t ≥ → f (t) ≥ f (1) = t − t − t + ≥ Second Solution (Mharchi Abdelmalek): LHS−RHS ≥ ⇔ cyc a3 2a2 − +a + b b cyc a2 − 2a + b b √ ≥0⇔ cyc a3 √ − a b + cyc (a − b)2 b ≥0 Which is Obviously true! Problem 96 ‘Hojoo Lee’ (Redwane Khyaoui): If a + b + c = then prouve that √ √ a b abc 3 + + ≤1+ a + bc b + ac c + ab Solution (Endrit Fejzullahu): Let x = ab ,y = c bc ,z = a ca We see that xy + yz + zx = ,and b a = zx, b = xy and c = yz Then we can write : (A,B,C are angles of an acute angled triangle) x = tan A B B C C A tan , y = tan tan , z = tan tan 2 2 2 52 A B C Then a = tan2 tan tan , and so on 2 After making the substitutions ,and using trigonometric transformations formulas ,the given inequality becomes : √ √ tan A2 3 3 + + ≤ + ⇐⇒ cos C + cos B + sin A ≤ + tan2 C2 + tan2 B2 + tan2 A2 √ C 3 cos C + cos B + sin C ≤ sin + sin C < ⇐⇒ 12x2 < 27 + 16x4 2 where x = sin C Problem 97 ‘—’ (Endrit Fejzullahu): If a1 , a2 , , an are positive real numbers such that a1 + a2 + + an = 1.Prove that n n ≥ − 2n − i=1 Solution (Mharchi Abdelmalek): Using The Cauchy Shwarz inequality we have n i=1 = − n i=1 (a1 + a2 + an )2 1 a2i ≥ = ≥ 2ai − 2(a1 + a2 + an ) − a2i − a2i 2− n = n 2n − Problem 98 ‘MOSP 2001’ (Mharchi Abdelmalek): Prove that if a, b, c > have product then : (a + b)(b + c)(c + a) ≥ 4(a + b + c − 1) Solution (Endrit Fejzullahu): Since abc = Then (a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) − ≥ 4(a + b + c − 1) ⇐⇒ ab + bc + ca + ≥4 a+b+c By the Am-Gm inequality ab + bc + ca + 3(ab + bc + ca)3 ≥44 a+b+c 27(a + b + c) It is enough to prove that 3(ab + bc + ca)3 ≥ 27(a + b + c)or(ab + bc + ca)3 ≥ 9(a + b + c) By AM-GM ,we know that (ab+bc+ca)2 ≥ 3abc(a+b+c) = 3(a+b+c) , so it suffices to prove ab+bc+ca ≥ ,which is obvious 53 Problem 99 ‘—’ (Endrit Fejzullahu): Let m, n be natural numbers Prove that sin2m x · cos2n y ≤ mn nn (m + n)m+n Solution (socrates): Use AM-GM (or weighted AM-GM) to get sin2 x cos2 x +n· = m n 2 2 sin x sin x sin x cos x cos x cos2 x + ≥ + + + + + m m m n n n = sin2 x + cos2 x = m · m times n times ≥ (m + n) 2 ( sinm x )m ( cosn x )n m+n etc Problem 100 ‘Mircea Lascu’ (Mircea Lascu): Let a, b, c > Prove that a b c + + ≥2 a b+c c Solution (Vo Quoc Ba Can): Write the inequality as c a b+c + + ≥3 a b+c c In this form, we can see immediately that it is a direct consequence of the AM-GM Inequality 54 ... that b2 c2 a2 + + ≥1 2 a + 2b b + 2c c + 2a2 Solution (Popa Alexandru): We start with a nice use of AM-GM : a2 a2 + 2ab2 − 2ab2 a(a + 2b2 ) 2ab2 2 = = − ≥a− a2 b2 2 2 a + 2b a + 2b a + 2b a + 2b... holds (a2 + 2) (b2 + 2) (c2 + 2) ≥ 3(a + b + c )2 Solution (Popa Alexandru): Cauchy-Schwartz gives : (a2 + 2) (b2 + 2) = (a2 + 1)(1 + b2 ) + a2 + b2 + ≥ (a + b )2 + (a + b )2 + = ((a + b )2 + 2) 2 And... (xy − 1 )2 ) ≥ 6x2 y ⇔ 4xy + (xy − 1 )2 · x + (xy − 1 )2 − 2x2 y ≥ We have that: 4xy + (xy − 1 )2 · x + (xy − 1 )2 − 2x2 y ≥ ≥ 4xy + 2( xy − 1 )2 − 2x2 y( because x ≥ 1) = 2x2 y + − 2x2 y = 2xy(y −