When do we have equalitỷ Gazeta Matematica Solution: First, we write the inequality in the following form (1+ 32) (+3) (+) (+Š) >7, But this follows immediately from Huygens Inequalitỵ We have equality for r=2y=s.2=1, 11 | Mihai Piticari, Dan Popescu | Prove that 5(ả +b? +0”) < 6(ả +b? +c?) +1, for all a,b,c > O witha+6+c=1 Solution:
Because ø + b + e = 1, we have a + b3 + cŸ = 3abe + a2 + b + cŸ — ab — be — cạ The inequality becomes
5(ả+b?+c?)_ <_ 18abe+ 6(a2 + bŸ + c?) — 6(ab + be + ca) + 1 ©
ÿ 18abe + 1 — 2(ab + be + ca) + 1 > 6(ab + bc + ca) <®>
ÿ 8(ab + be + ca) < 2 + 18abe © 4(ab + be + ca) < 1 + 9abe S (1 — 2a)(1T— 2)(1 — 2e) < abe ©
© (b+ec—a)(ce+a—b)(œ+b— e) < abe,
ÿ
which is equivalent to Schur’s Inequalitỵ
12 [| Mircea Lascu ] Let 71, 2%2, ,%, € R, n > 2 and a> 0 such that x, + 2 2 đa + +0„y = a and z7 +z2 + +22 < 2 T Prove that x; € 0, | tr all n i€ {1,2, ,n} Solution: Using the Cauchy-Schwarz Inequality, we get (ø—#i)” <(n— 1) (z3 + gã +: + #2) < (n— 1) (S-3): Thus, 2 2 2 2 2a
a” — 2ax, + aj <a*—(n-l)ti S 1% = — — <0
Trang 213 | Adrian Zahariuc | Prove that for any a,b,c € (1,2) the following inequality holds ba cvb a/c 4bVẽ cVa ` 4eJa—avb 4avb — bực 2h Solution: The fact that a,b,c € (1,2) makes all denominators positivẹ Then ba a fi ofa = aabte ©_ b(a+b+e)> vă4bVe—- ca) > ©_ (a+b)(b+c)> 4bVae,
the last one coming from a + b > 2Vab and 6+ c > 2Vbc Writing the other two inequalities and ađing them up give the desired result
14 For positive real numbers a,b,c such that abc < 1, prove that a b + + >a+b+ẹ boc oa First solution: If ab+ be +ca < a+6+c, then the Cauchy-Schwarz Inequality solves the problem: (x+b+e€) 1 1 1
a 6 ¢_ @etPateb athe Satbete
boc oa abc — abc —
Otherwise, the same inequality gives (ab + bc + ca)? oe _ Werbared, atbte Cc OG abc >atb+e 1S abe (here we have used the fact that abc < 1) Second solution:
Replacing a, b,c by ta, tb, te with t = = preserves the value of the quantity in the left-hand side of the inequality and increases the value of the right-hand side and makes at - bt - ct = abct? = 1 Hence we may assume without loss of generality that abc = 1 Then there exist positive real numbers 2, y, z such that a = x b= ¬ c= `,
The Rearrangement Inequality gives ° ' °
Trang 3‘Thus =øa+b+ec boc wv +ỷ423 ` z2 + 2z + z?œ ca 272 LYZ as desired Third solution: Using the AM-GM Inequality, we deduce that 2a b 3/ a" —T +~ >ở\| — > 3ạ Den be = 84 2b 2 Similarly, —+ < > 3b and —+ > 3c Ađing these three inequalities, the conclusion c a ạ is immediatẹ Forth solution:
fab* [cat | be*
Let # = \ y= ' w= a Consequently, a = xỷ,b = zx7,¢ = yz’, and also xyz < 1 Thus, using the Rearrangement Inequality, we find that
a z? ` oe ay 2
3>” >”n` "nh
15 [ Vasile Cirtoaje, Mircea Lascu | Let a,b,c,2,y,z be positive real numbers such thata+a>b+y>c+zanda+b+c=2+y+2z Prove that ay+bzr > ac+2z Solution: We have ayt+bx—ac—xz = ăy—c)+x(b—z) = ăa+b—x—-z)4+2(b-z) = ăa—x)+(a+2z)(b—z) = 1 s 1 1 » 1 = s(a—#)” + s (47 —#z”)+(œ+z)(b— z) = s(aT—#)” + s(a + #)(ø — # + 2b — 22) = 1 1
The equality occurs when a= 2, b=2,c=yand 2x >y+42z
Trang 4From (x + y +z)? > 3(ay + yz + zx) we get > 1 9 ————">~ + Ta" + z + zz (cty+z) so it suffices to prove that 9 1+ > 6 (c+ytz)? ~a+yt+z 3 2 The last inequality is equivalent to (1 — ==) > 0 and this ends the proof z++z 17 Let a,b,c be positive real numbers Prove that a3 bồ c aa b c2 JBMO 2002 Shortlist First solution: We have g3 2
ao > ta— beat +b > ablat) & (a—b)*(a +b) > 0,
which is clearly truẹ Writing the analogous inequalities and ađing them up gives eo bồ 8 @ b2 c2 2b? 2 —=+=š=+—>>_— > —b+—+ủ-— — —=—+—>+— B8 g2 4 TT cr 7 re-a bẹ a Second solution: By the Cauchy-Schwarz Inequality we have 3 3 3 2 2 2\ 2 a b C a b c b —4+—~+4+—}]>/[/—4+—4— (a+ +o (B+5+S)2(F +—+ ) : so we only have to prove that 2 p2 2 —=+T+>a+b+ẹ b Cc a
Trang 5and it is clear, because n > 3 and a; + aj41 + aj42 < a1 + ao +-+-+ a, for all 2
19 [| Marian Tetiva | Let 2, , z be positive real numbers satisfying the condition ả ty +27 4+ 2Qaeyz = 1 Prove that 1 a) 8Ụ2 € gì , b)z++z< 3 C) #U-++z +z <S— <#2 tụ?) + z9; ho | | Co d) sytaztyz < = 4+ 2zyz Solution: a) This is very simplẹ From the AM-GM Inequality, we have
Daả tỷ +2? + 2z > 422532) => aỷ2? < se —8
b) Clearly, we must have x,y,z € (0,1) If we put s = z+-+z, we get Immediately
from the given relation s?—284+1=2(1-2)(1-y)(1-2z) Then, again by the AM-GM Inequality (1 — x, 1 — y, 1 — < being positive), we obtain _ 1_ 1_z\Š xv fa asti<2(t xt ~- “| =:(“) After some easy calculations this yields 2s° + 9s? — 27 <0 © (28 — 3) (s +3)” <0
and the conclusion is plain
Trang 6On the other hand,
1 = a 4ỷ +274 2Qaeyz > Qayt 274 2cyz > > 2zy(1+z)<1—-z?>2zu<1-—z Now, we only have to multiply side by side the inequalities from above # +1 — 2# <€ 5 and z<l—2z to get the desired result: 1 1 LZ+ yz —2ryz< 5 — # © + + #z + Uz < 21222 It is not important in the proof, but also notice that 1 1 bby — dey = ay (242-2) > 0, 1 1 because — and — are both greater than 1 + Ụ Remark 1) One can obtain some other inequalities, using z+2z <1 and the two likes + 2zz < 1, z + 2z < 1 For example, multiplying these inequalities by z, y, x respectively and ađing the new inequalities, we get a tỷ +27 4+ b6eyz<x2+yt+z, or 1+ 4zz <S+z++z
2) If ABC is any triangle, the numbers
Trang 720 | Marius Olteanu | Let 21, 22,73, 74,25 € Rsothat x1) +%2+¢3+24+25 = 0 Prove that
| cos z;| + | cos zg| + | cos x3| + | cosx4|+|cosz5| > 1
Gazeta Matematica Solution:
It is immediate to prove that
|sin(z + ø)| < min{| eosz| + | cos y|,| sin z| + | sin y|}
and
| cos(x + y)| < min{| sin] + | cos y|,|sin y| + | cos z|}
Thus, we infer that
5 5
1 = |cos (>: 2] | < |coszi|+| sin (>: 2] | < | cos x1|+| cos r2|+] cos(xg+x44+25)|
But in the same manner we can prove that
| cos(ag + 24 + 25)| < | cosax3| + | cos x4] + | cos 25|
and the conclusion follows
Trang 8Second solution: Another improvement is as follows Start from 1 1 1 1 1 1 3.2 2.2 2.2 2.2.2 ep p?gtn tp let +#“z“ + “z4 “U27; which is equivalent to (+ #z + 9z)” > 2z (w + + z) ++22z? = 3( + +2)” Further on, (cy +ez+yz—3)? = (cyt az+yz) —6(ay+az+yz)+9> >3(++z)Ÿ—6(øu + zz + z)+9= 3(z2°+?+z°) +9, so that + + øz +z >> 3+ W3(z2+?+z2) +9 But 3(z2+2+ z2) +9 > Vz2+1+V2+1+Vz?~+1
Trang 94B—2A for any a,b,¢> 0 Let A= 2c-+b, B= 2a+e, C= 2b+ạ Then 2= ST SỐ —*S, A+4C —-2B B+4A-— b= ——^ c= ae and the inequality (1) is rewritten as C A PB BC HA “+C=~+~+4Í[ +<+<]|>1ư at Btat (2+5+5)>1 Because A, B,C > 0, we have from the AM-GM Inequality that CoA —-+L =1 —> i
and 7A + B + c3 3 and the conclusion follows
An alternative solution for (1) is by using the Cauchy-Schwarz Inequality: a b € ả b? c (a+b+c)? + + = + + > > 1 2c+b 2a+c 2b+a 2ac+ab_ 2ab+cb 2bc+ac — 3(ab + be + ca) 23 Let a,b,c > 0 with a+6+c= 1 Show that 2 2 2 a +b 0ð tee Fay b+e c+a a+b — Solution: Using the Cauchy-Schwarz Inequality, we find that 2 a+b (Soả +1)
b+e ~ SP(b+ơ Soả + So ab
And so it is enough to prove that
wy + `ab 281 (Se) > 2S ăb+ 0) +2) ab
Trang 1024 Let a,b,c > 0 such that at + 64 + ct < 2(ảb? + bc? + c?a”) Prove that ả +b? +c? < 2(ab+ be + ca) Kvant, 1988 Solution: The condition So at < 2S `ả0? 1s equivalent to (a+b+c)(at+b—c)(b+c—a)(c+a—6b)>0 In any of the casesa = 6+ c, b=c+a, c=a+b, the inequality Soa < 2S ab
is clear So, supposea+bf#¢c, b+cH#a, c+aF b Because at most one of the numbers 6+ c—a, c+a—b, a+b-—c is negative and their product is nonnegative, all of them are positivẹ Thus, we may assume that
a* <ab+ac, 6? < be+ba, 2 <ca+cb
and the conclusion follows
Trang 1126 | Marian Tetiva | Consider positive real numbers x, y, z so that #2 + g7 +z?= 112 Prove the following inequalities a) xyz > 27; b) z + zz + 0z > 2ï; c)z++z>9; đ) z +zz+z >> 2(z++z)+9 Solution:
a), b), c) Using well-known inequalities, we have
ays = 2? +ỷ +2? > 38/22y222 = (xyz)? > 27 (xyz),
which yields xyz > 27 Then cy +02 +yz > 34) (xyz)? > 3V272 = 27 and z-++z>32zwz > 327 =9 d) We notice that x27 < xyz > x < yz and the likes Consequently, 2 < z-øz => 1< z2 => z >1 Hence all the three numbers must be greater than 1 Set œ=#—l,b=p—l,c=z—] We then have a > 0,6 > 0,c¢> 0 and #œ=a+l,=b+l1,z=c+Ị Replacing these in the given condition we get
(a+ 1)? + (641)? + (e411) = (at+1) (64+ 1) (c+)
which is the same as
a2 + b°® +c?2+a+b+ec+2= abẽ+ ab + ae + bẹ If we put g = ab+ ac+ bc, we have
q<a2+b?°+c°,/3qg<a+b+ec
Trang 12‘Thus q+ V3qt2 < a ®+b?+c°+a+b+c+2= 3 3 = abe-+ ab + ac+ be < 30) +ự and setting # = 4⁄3g, we have 3 z+2<;- ®(œ= 6) (ø + 3)” >0 Einally, V3q = z > 6> q= ab + øe + be > 12 Now, recall that a=a2—1,b=y-—1,c=z-1, so we get (ứZ—1)(@—1) + (z—1)Œ@_—1I)+(g—1)(z-—1)>12> > z+zz+z>2(z++z)+9;
and we are donẹ
One can also prove the stronger inequality + + øz + z 4(x++z)—9 Try it! 27 Let x,y,z be positive real numbers with sum 3 Prove that V2 + WW + Vz 3 #ụ + z + zz Russia, 2002 Solution: Rewrite the inequality in the form e+Watyt2y+242/z>x? +y +22 4+2Qry+Qyzt2zrs ®#z”+2Vz +ˆ+2Vy+z”+2vz > 9 Now, from the AM-GM Inequality, we have z2 + 2V = #2 + V2 + VW# > 322 - z = 3z, ? + 2W > 3, z? +2Vz > 3z, hence z2 +ự + z2 + 2(V# + vW + v2) >3(z+„+z)>9
28 [ D Olteanu | Let a,b,c be positive real numbers Prove that
atb a ote b cota c 3
b+e 2a+b+e cta 2%+e+a a+b 2c+a+b— 4
Trang 13Solution:
We set x =a+b,y=b+cand z=a+cand after a few computations we obtain the equivalent form
#1 2 x y Z 9
—+—+—+ + + > nx
Yo o£ 4# “e£+Yy YyYre z+z 2
But using the Cauchy-Schwarz Inequality, x z x Z z++z)2 z++2)? 2U 2 ye 5 fetytey _ (@+y :) —= UJ 2 #é #190 Ye Z2+2 #Ặ+Uz +22 + +z z2 +24 T+ụˆ+z 2+ „+ z)! ` 8 ++z)° 2(z + z + z#)(œU + Uz + z# + z2 Ð12+ 2z?) — (xụưz+zz+(x++ z)®)Ÿ
and the conclusion follows
29 For any positive real numbers a, b,c show that the following inequality holds a b —~+-+-> b c a ẹct+a e+tb a+b ate b+a b+e India, 2002 Solution: ; a Cc Let us take = #,— =1,— = z Observe that C a ate l+zụ l—# ——=———=z++ b+e l+%y 1+ Using similar relations, the problem reduces to proving that if ryz = 1, then zx-l y-1 2-1 + + >0 © +1 zZT+1 z+1” © (z ”—1)(z+1)+(g?—1)(z+1)+(z?—1)(y+1)>0 © - SN z?z+ sả > Soa t3
But this inequality is very easỵ Indeed, using the AM-GM Inequality we have » xz > 3 and so it remains to prove that » x? > » x, which follows from the
inequalities ›
» x > 3) > » 2
30 Let a,b,c be positive real numbers Prove that
a3 ~ b3 ~ c ` 3(ab + be + ca)
b2—bc+c2 c2-ac+a2 a*—-ab+h 7 atbte
Proposed for the Balkan Mathematical Olympiad
Trang 14
First solution: Sinceat+b+c> 3(ab + be + ca) at+bt+e ả + bŠ + c5 b2—bc+c2 c2—-ca+ò2 a*®—ab+b? From the Cauchy-Schwartz Inequality, we get
ĐT b2 — be +c? ~ So al(b? — be+ (Sả) ec?)
Thus we have to show that
, it suffices to prove that:
>~a+b+c
(a” + ðb2+ c2)? >(a+b+e) So alẻ — be +c’)
This inequality is equivalent to
at +b' 4c! + abela + b+.) > ablả + b2) + bc(b? +c?) + calc? + a”),
which is just Schur’s Inequalitỵ Remark The inequality ả b3 c b2 — be + c2 T c2 — ca + a2 T ả — ab + b? was proposed by Vasile Cartoaje in Gazeta Matematica as a special case (n = 3) >atb+e
of the more general inequality
2a" —b" —c™— 2b” — ce” — a” 2c — at — OF Sg b3 — be + c2 c2 — ca -+ q2 a2—qœb+ b2 — ` Second solution: Rewrite our inequality as ` (b + e)a3 ` 3(ab + be + ca) B+e ~ a+b+c `
Trang 15and _ B+ea | (b+c)(b2T— be+c?) — B=) Rohe 2 Boke 2 So we get 1 A+B = (Se)(Seaae)- = s 0+0? —be+€)) (= mì > ; (> vơe) from the Cauchy-Schwarz Inequalitỵ Hence A>s(X v12) -2)a=S)va+b-vi+e- 2a We denote ab A= Veen Vireo = eh VÀ ab+a° +a 2 and the similar relations with Ay and A, So A > A,+A,+A,- But because » a) > 3 (> ab) we get 3 `ab _ 3 dab (Sa) (Xe) 3 +a-+a 2 (XH and also the similar inequalities are truẹ So we only need to prove that A, > 2 1 +a*—-a >a+b+tceœ ` +(e) >2 J1
We consider the convex function f(t) = 3 +t? Using Jensen’s Inequality
we finally deduce that
1 a °
— 5+ (aie) > ———— | >2
We have equality if and only ifa = b= cẹ
31 [ Adrian Zahariuc | Consider the pairwise distinct integers 71, %2, ,2n, n > 0 Prove that
Trang 16Solution: The inequality can be rewritten as nm So (ai —#¡+1)Ÿ > 2(2n — 3) 21
Let #„„ = min{Z1,#a, ,#„} and #ạ;¿ = max{z41,#a, ,#„} ŠSuppose without loss of generality that m < M Let
Si, = (Lm — #m+1)Ÿ + -+ (ZAZ—1 — au)? and So = (tm — @M41)° + -+ (Ln — a1)? + ti — ze)" + -+ (Lm—1 — Lm)* k 2 k 1 The inequality » a; > i (>: «| (which follows from the Cauchy-Schwarz In- z=1 +1 equality) implies that (#w — #m)Ÿ Ss; > '=""“M—m and 2 LM —XLm > ————— S22 (i —m) So * 2 2 1 1 2 „(8i — #rm) = 5+ 52 Đ (am — #m) (r-=†tz=mr=m)È 4 4 > (n— 1) — =4n—-Ñ+— >á4n-— 8 n n But a a So (ai — a1) = »- — #¿11 — 0 mod 2 i=1 i=1 so nm » —#¿+1)Ÿ > 4n — 6 21 and the problem is solved
Trang 17for all 41,2%2, ,%n—1,2%n > O which ađ up to 1 Let us prove first the inductive step Suppose the inequality is true for n and we will prove it for n + 1 We can of course assume that x2 = min{2%1,2%2, ,£%n41} But this implies that
2 2 2 2 2 2 2
Ujtg + egag t+ +a7a < (a1 +22)" 43 + 45044 °+°4+ 05 1 on + U7, (21 + 22)
But from the inductive hypothesis we have
4
(đi +22) #3 + 2314 + + +12 18a + #2 (#1 + #a) < >
and the inductive step is proved Thus, it remains to prove that a2b + bŸc + ca < 4
ifa+b+c=1 We may of course assume that a is the greatest among a, b,c In this case the inequality ab + b2e + ca < (a + a (0 + 5) follows immediately from abe > 67c, ae > ac Because +; a+ a = a = 1=—2 2 2 2 4 we have proved that a2b+ b?e+ c?a < 7 and this shows that the maximum is indeed 4 27
33 Find the maximum value of the constant c such that for any Z1,#2, ,p,''' >0 for which xp41 > 2, + 4% + -+ a, for any k, the inequality
V1 + V#s + - + VW#n < €V#1 + øa + : +2
also holds for any n
IMO Shortlist, 1986 Solution:
First, let us see what happens if zz¿¡ and #1 -Ƒ#a +- +#¿ are close for any k For
example, we can take x, = 2", because in this case we have x1 +aơ: +a,% = ©p41—2 Thus, we find that n › v2 k=1 c>
for any n Taking the limit, we find that c > 1+ V2 Now, let us prove that 1+ /2 works We will prove the inequality
Trang 18by induction For n = 1 or n = 2 it is clear Suppose it is true for n and we will
prove that v/#1 + 4⁄22 + : +4/#„ T /#a+i < (1 + ⁄2)⁄21 +#a + ': +#np T#an+1
Of course, it is enough to prove that Vtnt+1 < (14+ v2) (2i + #2 T-'* †#n‡L— V41 +#a +: C#n) which 1s equivalent to V?\ +8 + +** 8n + Vi +8 + tem < (14 V2) Vent But this one follows because #1 ta T - Ð+đa € #n+1 34 Given are positive real numbers a,b,c and x,y,z, for which a+ a2 =b+y= c+2z=1 Prove that 1 1 1 (abe + xyz) (— +—+ —) > 3 ay bz cx Russia, 2002 Solution:
Let us observe that abc + xyz = (1 — 6)(1—c) + ac+ab—ạ Thus,
l-e a T 1—b c pn abc + xyz ă1 — b) Using these identities we deduce immediately that
3+ (rye +abe) (“+ b+ 2) = a, 8 5 ¢ toe tra tn) ay bz c@ l-c l-a 1-6 a b Cc
Trang 19Second solution:
Because the inequality is homogeneous we can consider without loss of generality that a+ 6+c¢=1 and so the inequality is equivalent to 1 1 3 ăœ + 1) — 4abe 1 1 We have =—_— , So the inequality is equivalent to t t(t+1) t+1 1 1 1 —< 2a < anit tube We will prove now the following intercalation: 1 9 1 1 1 -<- < ———
2a — 1+ Tabs < Gai + Tobe
The inequality in the right follows from the Cauchy-Schwarz Inequality: H) (X«+1)) >9 and the identity (a + 1) =4 The inequality in the left can be written as » ab< 1+ 9abe 4
, which is exactly Schur’s Inequalitỵ 36 Find the maximum value of the expression
a”(b+ec+đd) +~b?(e+d+a) +c(d+a+b) + dđ”(a+b+ e) where a,b,c,d are real numbers whose sum of squares is 1
Solution:
The idea is to observe that ả(6+ce+d)+6?(c+d+a)+c?(d+a+b)+d?(at+b+c) is equal to » ab(a” +b*) Now, because the expression ab(ả +6?) appears when writing
(a — b)*, let us see how the initial expression can be written: 44 54 +4 6a7b? — (a — b)* bả+?) = SoS = Sabla + 6°) » 1 3À 2a2+6S ả?— À (a -b)* — 3—š(a—b) 4 ˆ 4 4 <2 — 4 1
The maximum is attained fora =b=c=d= 5"
37 | Walther Janous | Let x,y,z be positive real numbers Prove that
x y Z <1
>+V@dtwGa+z) w+Vp+202s) z+V+sap—
Trang 20First solution: We have (a+y)(x+z) = ryt (ả +yz) +22 > zư2z WWzZ+zz = (V#yW+ v2) Hence an (x + y)(x + 2) SU aa + _ Vax at ym VE Vấn But
Jet jit feo
and this solves the problem
Second solution:
From Huygens Inequality we have 4/( + y)(x + z) > «+ ,/yz and using this
inequality for the similar ones we get % Z CỐ Ù r+ V000) yt+Vưaưt) z1v0isb am x y Z + + 2e+/yz Wt VSze 2z+zụ V⁄z b— Vzz oe NI £ Ụ 2 z 1 + 1 + 1 <1 2+a 24+6b 2+c—
From the above notations we can see that abc = 1, so the last inequality becomes after clearing the denominators ab + bc + ca > 3, which follows from the AM-GM Inequalitỵ Now, we denote with a = and the inequality becomes 38 Suppose that aj < ag < < Gy are real numbers for some integer n > 2 Prove that 4 n° 4 ai dỆ + A903 + + Anat > a2a4 + a3Q3 + +4a4)4 Iran, 1999 Solution:
A quick look shows that as soon as we prove the inequality for n = 3, it will be proved by induction for larger n Thus, we must prove that for any a < b < c we have
ab(b® — a®) + be(c? — 6?) > căc? — a’) & (c? — b°)(ac — be) < (b® — a®)(ab — ac) Because a < b < c, the last inequality reduces to ăb? + ab + ả) < e(c2 + be + b3)
Trang 21Solution: 1 1 ' Using the inequality ——— < — + — we infer that x+y” 4z Ay 4a a a Ab bb Ac <-+~-, <-+- and b+c—b c` qa+c a Ce a+b Ađing up these three inequalities, the conclusion follows <£.¢ —~a b
AỌ Let @1,@2, ,@, > 1 be positive integers Prove that at least one of the numbers %/a2, %%/a3, , °-W/@n, *%/a1 is less than or equal to V3
Adapted after a well-known problem
Solution: ,
Suppose we have đc) > 33 for all ị First, we will prove that nn < 33 for all natural number n For n = 1, 2,3,4 it is clear Suppose the inequality is true for n > 3 and let us prove it for n + 1 This follows from the fact that m + Ì 1 1 Cn : 1+—<1+7< W333 = 3-38 > n=n-+1 nr 1 1
Thus, using this observation, we find that a; > 33 > Q;4, => Gi41 > a; for all 2,
which means that a, < ag <-+- < Gn_1 < Gy < a1, contradiction
41 | Mircea Lascu, Marian Tetiva | Let x,y,z be positive real numbers which
satisfy the condition # + #z + 9z + 2zz = 1 Prove that the following inequalities hold 1 < —: a) yz S 3 , b)aty+z2 5; _- + +2); ta Ụ Z _~ PU a); 111 22-1) d) Fee Aleut) > EIT, where «= max {2,52} Solution: a) We put t? = xyz; according to the AM-GM Inequality we have 1=z+zz+z+ 2zuz > 32 + 2 @ (2E— 1) (+ 1)ˆ <0,
therefore 2ý — l <0 ©< > this meaning that ryz < ¬
b) Denote also s = # + + z; the following inequalities are well-known
(z++z)Š>3(zu + zz + 0z)
and
Trang 22then we have 2s” > 54xyz = 27 — 27 (ay + xz + yz) > 27 — 9s”, 1 ẹ 2s3 + 9s? — 27 >0 © (2s — 3) (s+ 3)” >0, where from 2s—3>0<©s> 5: 1 Or, because p < 3? we have 2 9 27> 3¢=3(1-2p)>3(1 )=-; if we put g= xy t+auz+ yz, p= xyz
Now, one can see the following is also true 3
q= + #Z + W2 3 1:
1
c) The three numbers 2, y, z cannot be all less than 5 because, in this case we get the contradiction
3 1
% + zz + Uz + 2xz < 11218 = ]; 1 because of symmetry we may assume then that z > =
We have 1 = (224+ 1)ay+2z(a@+y) > (2241) ay + 2z,/ey, which can also be written in the form ((2z + 1) /zy— 1) (,/ey+1) < 0; and this one yields the inequality 1 4 < ———s (2z + 1)” 2
Trang 231
(the assumption z > — allows the multiplication of the inequalities side by side), and this means that the problem would be solved if we proved 2(224+3) 2241 2 (22 +3) ` 221 D2163 42+ 4z 41: 2z+1 Z 1 but this follows for z > 2 and we are donẹ 1 d) Of course, if z is the greatest from the numbers #,#,z, then z > 3) We saw that 1 1 1 cty 4+0) = (ety) (S-4)> AF eee 2(2z — 1) (2z +3) 1 (2z-1)° aan ——“'=4¿ +-^ 2z4+1 z 2(2z4+1) ——, where from we get our last inequality 1 11 (2z —1)° —=+—=+—-=4 > ees x2 (t+y+2) 2 ay
Of course, in the right-hand side z could be replaced by any of the three numbers which is > 5 (two such numbers could be, surely there is one)
Remark
It is easy to see that the given condition implies the existence of positive numbers b
a,b,c such that « = ——,% = ——,z = — And now a),b) and c) reduce
b+e c+a a+b
immediately to well-known inequalities! Try to prove using this substitution d) 42 | Manlio Marangelli | Prove that for any positive real numbers 2, y, z,
3(z2 + ỷz + 272) (aỷ + yz? + za”) > xụz(œ + + z) Solution: Using the AM-GM Inequality, we find that 1 yz ry” SY ÿ/22 —+— - —+ >
3 0 2z+z2z+z?u z?+zz? + arỷ ~ 3-(u3z + 22a + xy) - (yz? + zx? + aỷ)
and two other similar relations 1 + z”z + ụz? ` 3z #/zz 3 1 2z+2z2z+z”u 22+zz”+zụ? — ÿ/3-(u2z + z?3z + +?) - (wz2 + zz2 + xỷ) 1 z2 zu 3# ở/2U2 —+— - —+ >
3 0 2z+z2z+z?u z?+zz? + arỷ ~ 3-(u3z + 22a + xy) - (yz? + zx? + aỷ)
Trang 2443 [| Gabriel Dospinescu |] Prove that if a,b,c are real numbers such that
max{a, b,c} — min{a, b,c} < 1, then
1+a@?4+b% +3 + 6abe > 3a7b + 3bŠc + 3c2a Solution:
Clearly, we may assume that a = min{a, b,c} and let us writeb =a+a2,c=a-+y,
where x, y € [0, 1] It is easy to see that ả +6? + c? —3abe = 3ă#?T— zử)+z3 +
and a7b 4+ b?c + ca — 3abe = ăx? — xy + y*) + x?ỵ So, the inequality becomes
1l+2?+ỷ > 3x7ỵ But this follows from the fact that 1 + z3 + > 3z > 3z”, because 0 < #, < 1 44 | Gabriel Dospinescu | Prove that for any positive real numbers a, b, c we have ? b2 ? 1 1 1 o74+ (2+) (24—) (24+ —)>6+b4+0(-+=4-) be Ca ab a boe Solution:
By expanding the two sides, the inequality is equivalent to 2abe(ả + b3 + c3 + 3abc — ảb — ảc — b?a — b?e — c?a — 7b) + (ảb3 + b?c3 + ca3 + 3ảb)c? — a”b?e —
a’ bc? — abc? — ab’? — ảb?c — a’ bc?) > 0 But this is true from Schur’s Inequality applied for a,b,c and ab, bc, cạ 1 az 1 45 Let ap = 2 and ø;+¡ = øy + a Prove that 1- — <a, <1 TST Singapore Solution: 1 1 We have =—- and so Ak+1 Gk đụ + Tì Xa) “mm k=0 a k+1 1 7 <b ee <bean <Ị an Now, because the sequence is increasing we also have 2 — — = ` Gn k= 6 2 + aK > ai and from this inequality and the previous one we conclude immediately that 1 l——<ø, < ] n
46 | Calin Popa | Let a,b,c be positive real numbers, with a,b,c € (0,1) such
that ab+ be + ca = 1 Prove that
a b C ‘(= 1-6? —)
+ 5
Ss 2
Trang 25Solution: It is known that in every triangle ABC the following identity holds A B A » tan 5 tan 2“ 1 and because tan is bijective on |0 3) we can set a = tan > b= B C
tan 377 c= tan — The condition a,b,c € (0,1) tells us that the triangle ABC is acute-angled With this substitution, the inequality becomes A „ A 2 tan > 1 — tan 3 1 5, ——aÈ3 ———a 9®) tan4>35 9 1 — tan 5 2tan
& tan Atan Btan C(tan A+tan B+tan C) > 3(tan Atan B+tan BtanC+tanC tan A) S
& (tanA+tanB+tanC)? > 3(tan Atan B + tan BtanC + tanC tan A), clearly true because tan A,tan B,tanC > 0
47 | Titu Andreescu, Gabriel Dospinescu | Let z,,z < 1 and z++z = 1 Prove that 1 1 1 < 27 1+2 13g 1+2 T10 Solution: Using the fact that (4 — 3t)(1 — 3t)? > 0 for any t < 1, we find that 1 27 < —(2—3) Tea 2 59?-*) Writing two similar expressions for y and z and ađing them up, we find the desired inequalitỵ Remark Tough it may seem too easy, this problem helps us to prove the following difficult inequality ` (b+e— a)? ` 3 a2+(b+c)? — 5
In fact, this problem is equivalent to that difficult onẹ Try to prove this!
48 | Gabriel Dospinescu |] Prove that if /a + ⁄# + ⁄z = 1, then
(—=z)?1—ø)”q~ 2z) > 2'zwz(z + w)(w + 2)(z +2)
Solution:
We put a = /a, b= fy andc= Vz Thenl—aw#=1-a@ =(a+b4+c7-v= (b+ c)(2a + b+ c) Now we have to prove that ((a + 6)(6+ c)(ce+ a)(Qa+b4+c)(at
Trang 264
{a+ oy (this being 8(b+c)(c+a)(a+b) as it follows from the following true inequalities ab(ả + 67) <
equivalent to (a—b)* > 0) and (2a+b+c)(a+2b+c¢)(a+b+2c) >
AQ Let x,y, z be positive real numbers such that xyz = x+y+2+2 Prove that (1) zƯz+zz>2(¿+ư2); (2) vz+ + vz< a Solution: The initial condition xyz = x +y+ 2+ 2 can be rewritten as follows 1 + 1 + 1 1 Ite 1+y l+z —` Now, let 1 1 1 =a, = b, —C l+z l+% l+z Then l-a Ob+e cta a+b rz = 5 = 5 = a a b Cc (1) We have b+e et+a eta a+b a+b b+e + + > b b c e a b 4ot ) oobi a8 + 3abe > ê zU-+OUz+zz > 2(z+ưz)<© > — a b > ab(a + b) + be(b + e) + căe + a) ` ăœ — b)(œ — €) >0, (= c+a 2 +: which is exactly Schur’s Inequalitỵ (2) Here we have 3 vz+vƯvz<s 22 oy a b tử b co C a3
b+e cta eta a+b œ+b b+ec—2'
This can be proved by ađing the inequality a b <_ 1 a 4 b b+ce c+ta” 2\at+e bt+e with the analogous ones
50 Prove that if x,y,z are real numbers such that x? + ỷ + z? = 2, then
z# + +z< zựz + 2
Trang 27First solution: If one of x,y, z is negative, let us say x then 2+z0z—#—U—z=(2—-—z)—z(l1— z)> 0, z2 +? because # + z < 4⁄2(2 + z?) < 2 and z < < 1 So we may assume that O<a<y<z.Ifz<1then 24+ayz-ex-y-z=(1-z)(1l-zy)+(-2z)(1-y)>0 Now, if z > 1 we have
zt(@t+y) < V22 + (+ +)”) = 2/1 + ry < 24+ ary < 2+ xyz
This ends the proof
Second solution:
Using the Cauchy-Schwarz Inequality, we find that
atyte—syz=2(l—yz)tytz< V2? + (y4+z)?)-(1+ (1 — yz)) So, it is enough to prove that this last quantity is at most 2, which is equivalent to the inequality (2 + 2yz)(2 — 2yz + (yz)?) <4 = 2(yz)? < 2(yz)?, which is clearly
true, because 2 > ỷ + z? > 2z
Trang 29But using the Cauchy-Schwarz Inequality and the AM-GM Inequality, we deduce that > Qj ~~ m—— -Ắ- m + /@a + (/6G;_—1 + v/G¿+1 TÐ + VJữn _ 2 — 7 Vm — Ì- Sai viên Van” Jas V8 ván Van Sees 7 ¿=1 + Qtr + đ—1 + 6+1 Ð +! + Vdn — nr a; > n—1 Z » —— 2+1 and we are donẹ
53 [ Titu Andreescu |] Let n > 3 and a1,a2, ,@, be real numbers such that
đi +aa+ +ø„ > n and ả+a2+ +aø2 > n2 Prove that rna#ø{a,da, , a„} > 9
USAMO, 1999
Solution:
The most natural idea is to suppose that a; < 2 for all i and to substitute
Trang 3054 | Vasile Cirtoaje | If a,b, c,d are positive real numbers, then a-6 b-c c~d d~-ay, b+e c+d d+a a+b—— Solution: We have a—b b+c ~ c+d b-c ~ d+a c-d ~ d-a a+b = a+c b+e b+d c+đ+d dt+a ct+ta d+b a+b -4= 1 1 1 1 — a — — 4 (a+©(pxzt1raz)} 0+9 (t3) Since 1 + 1 ` 4 1 + 1 ` 4 b+c_ d+a— (b+c)+(d+a) c+d a+b— (c+d)+(a+b) we get
a—b b+c + c+d b-c + c-d _ d-a dt+a + a+b~ > (b+e)4+(d+a) 4(ø + €) (c+d)4+(a+b) 4(b + d) —4=0
Equality holds for a = c and } = d
Conjecture (Vasile Cirtoaje)
If a,b,c, d,e are positive real numbers, then
a-6 b-c c~d d~e ẽay,
b+e c+d dt+e eta at+b7~ —
55 If x and y are positive real numbers, show that #Ơ + yđ > 1 France, 1996 Solution: a We will prove that ả > —————_ prov — a+b—ab
Inequality it follows that a!~’ = (1+a@—-1)!~? <14+(a—1)(1-6) =a+b-—ab
and thus the conclusion Now, it x or y is at least 1, we are donẹ Otherwise, let 0 < z,y < 1 In this case we apply the above observation and find that #¥ + y® >
Ụ > t + Yoo = 1
%+U—75U ˆ Lry—szwy 2+1 #+9
for any ø,b € (0,1) Indeed, from Bernoulli
56 Prove that if a,b,c > 0 have product 1, then
(a+ b)(6b+c)\(e+a) > 4(a+b+c-1)