First solution:
Using the identity (a + 6)(6+ c)(c +a) = (a+6+4+c)(ab+ be + ca) — 1 we reduce
the problem to the following one ab + be + ca + — > — > 4 atb+e Now, we can apply the AM-GM Inequality in the following form ,/ (ab + be + ca)? b+6 an ————— ——— 9(a+b+)
And so it is enough to prove that
(ab + be + ca)? > 9(at+b+c)
But this is easy, because we clearly have ab + be + ca > 3 and (ab+ be + ca)? > 3abc(a + b + e) = 3(a+b+ ©) Second solution: 3 We will use the fact that (a + b)(b+ c)(e +a) > gia + b+e)(ab + be + ca) 2 So, it is enough to prove that 9 (0b + be + ca) + > 1 Using the AM-GM ; a+b+c Inequality, we can write „| (œb + be + ca)? ————— >1 a+b+c7— 8l(a+b+ec) — ? 2 g (ab + be + ca) + because
(ab + be + ca)2 > 3abe(a + b + e) = 3(a+ b+ e)
57 Prove that for any a,b,c > 0,
(a7 + b` + c?)(a+b— e)(b+e— a)(e+a— b) < abe(ab + be + ca)
Solution:
Trang 258 | D.P.Mavlo | Let a,b,c > 0 Prove that 1 161 1)(b + 1)(c + 1 Btatbtesn4 rere Sal fy gle We Mer) b b Mes Dery) 1 + abe Kvant, 1988 Solution: The inequality is equivalent to the following one La Dato — _>939=————“^= eatery — abe + 1 or 1 2 a
abe Ya+ — +) a c+ 5 > 2(Soa+ am) But this follows from the inequalities 2 b 2 C 2 a œ“bc + P > 2ab, “ca + — > 2bc,c“ab + ọ > 2ca a and 5 1 2 1 9 1 a-c+ — > 2a,b°a+ — > 2b,c°b+ = > 2c e a b
Trang 3Of course, this is true for any other variable, so we can add all these inequalities to obtain that
which is the desired inequality
60 Let a,b,c,d > 0 such that ø + b+ e= 1 Prove that 1 1 a® +6° + + abed > min { 5, stat: Kvant, 1993 Solution: 1 Suppose the inequality is false Then we have, taking into account that abc < 2m 1 1 1 the inequality d (5 — abe) >@+P43 - 3 We may assume that abe < 7 Now, 1 we will reach a contradiction proving that a® + 6? + c? + abed > T It is sufficient to prove that 1 a+h+e—— 1 ———T—————— aửe +a3+b”+c7> + 7 abc But this inequality is equivalent to + l5abc > 1 We use now the identity 1+ 9ab » a3 = 3abe+1—3 ` ab and reduce the problem to proving that » ab< _ which is Schur’s Inequality
61 Prove that for any real numbers a, b,c we have the inequality
3 +22)? +b2)?(ø— e)?(b— ø)” > (1+a”)(1+ð2)(1+c”)(a— b)”(b— e)”(e— a)°
AMM
Solution: 5 52
(1 1
The inequality can be also written as » ren? assume that a,b,c are distinct) Now, adding the inequalities
(1+ a?)(1 + 07) (+6)d+e) J, 14+ 06?
(1+c?)(a—b)2 — (1+a@?)(6-c)? ~ |a— b|le— Ù|
(which can be found using the AM-GM Inequality) we deduce that (1+ a7)(1+ 0?) 14+ 6?
2>axzza- >> I6-s0-a
Trang 4and so it is enough to prove that the last quantity is at least 1 But it follows from » 1+? jy SS 1 + - |l(b— a)(b—e)|— —c) and the problem is solved = 1
Trang 5— — = (e+w+z~3) > “—(3ÿwz— 3) =0 Equality holds for # = = z = 1 Remark:
Using the substitution 6 = a+1(6> 2) andz= -, y= Đ z= - (abc = 1) the
inequality becomes as follows
1 1 1 3
a? (b +c) 7 b2(e + a) 7 đ*{(a + b) 2 2
For 6 = 3, we obtain one of the problems of IMO 1995 (proposed by Russia) 63 Prove that for any real numbers 21, ,2%n,Y1, -,Yn Such that 2?+ -+22 = yee dyad, k=1 ?ì (Z1a — #s1)” < 3 ( — > een Korea, 2001 Solution: We clearly have the inequality +? hr n 2
(amram YS wan? = (Set) (Sout)- (Sam) = 1<i<j<n i=1 i=1 i=1
- (1- Soa) mm) i=1 i=1
Because we also have mới <1, we find immediately that i=1 ( — 3 su) ụ + year] <2 ( — ` ea] i=1 =1 z—=1 and the problem 1s solved
64 [ Laurentiu Panaitopol | Let ai, ø›, , „ be pairwise distinct positive inte- gers Prove that
2n+1
aj tag to +a, 2S (ai + as + - + an)
Trang 6Solution:
Without loss of generality, we may assume that a, < ag < - < a, and hence a; > 17 for all 7 Thus, we may take b; = a; —i > 0 and the inequality becomes nr nr n 9 < ; mín + 1)(2n + 1) _ 2n +1 > : ; —————————, n(n + 1)(2n 4+ 1) ` bj +25 ib; + 6 >—a › Ù¡ + 6 i=1 i=1 +1
Now, using the fact that aj, > a; we infer that 6) < ¿ < - < ð„ and from Chebyshev’s Inequality we deduce that w=1 +1 7=1 and the conclusion is immediate Also, from the above relations we can see imme- diately that we have equality if and only if a,,a2, ,a@, is a permutation of the numbers 1,2, ,n 65 [ Calin Popa ] Let a,b,c be positive real numbers such that a+b+c¢= 1 Prove that b/c + ca + avb ` 3v3 a(V3c+vab) b(V3a+vbe) c(V3b+ ca) — 4 - Solution: Rewrite the inequality in the form bc > 3v3 ca + 4 ——+a b ¬ be ca ab With the substitution « = 4/—,y = „—'®=Ứ= the conditlon a + b-+e= 1 a c becomes xy + yz + zx = 1 and the inequality turns into > 3v “4 # 2 V3y + yz But, by applying the Cauchy-Schwarz Inequality we obtain 2 a » 7 ) 3 ty 3v3 » V3ay + ayz > V34 3xryz > V3 + i =—, 4 v3
where we used the inequalities
Trang 766 | Titu Andreescu, Gabriel Dospinescu | Let a, 6, c,d be real numbers such that
(1 + a?)(1 + 67)(1 +c?)(1 + d?) = 16 Prove that
—3 < ab+bce+cd+da+ac+ bd — abcd < 5
Solution:
Let us write the condition in the form 16 = H¿ +a)- Ta —i) Using symmetric
sums, we can write this as follows
16= (1-10 a- Sab +i > abe + abcd) (1+% 0 a- Sabi D abe + abcd)
So, we have the identity 16 = (1 — Ð` ab + abcd)? + (SS a— >> abc)” This means that |1 — S> ab+ abcd| < 4 and from here the conclusion follows
67 Prove that
(a? + 2)(b + 2)(c? + 2) > 9(ab + be + ca)
for any positive real numbers a, 6, c
APMO, 2004 First solution:
We will prove even more: (a? + 2)(Ù2 + 2)(c? +2) > 3(œ + b + c)? Because (x+b+e) < (la| + |b| + |c|)”, we may assume that a,b,c are nonnegative We will use the fact that if x and y have the same sign then (1+ 2)(1+y) >1+2+y So,
we write the inequality in the form
(+2) > othe
and we have three cases 2
2_ 4 2_ 4 ( a)
i) If a,b,c are at least 1, then TT (“ 3 +1) >14505 5 » 2)
Trang 8Second solution:
Expanding everything, we reduce the problem to proving that
(abc)? + 23 `a?0? +45 0a? +8> 9S — ab
Because 3300? > 3 `ab and 23 `a?0? +6> 4À ` ab, we are left with the inequality (abc)? + » a?+2> 2À ` ab Of course, we can assume that a,b,c are non-negative and we can write a = 27,b = y’,c = z” In this case
2S ab-S oa? =(z++z)(z+— z)(u+z— z)(z+z— 9)
It is clear that if x,y,z are not side lengths of a triangle, then the inequality is trivial Otherwise, we can take x =u+v,y =v+w,z = w-+u and reduce the inequality to
((u+v)(v+w)(w+u))*+2> 16( + 0 + u)uU
We have ((u+v)(v + w)(w +u))* +141>3%/(ut vs (vtw)t(utw)! and
it remains to prove that the last quantity is at least 16(u+v+w)uvw This comes
down to 3
16
(utv)i(v+w)*(w+u)* > sp (wap) (u ++u)
But this follows from the known inequalities 8 (utv)(vt+w)(w+u) > gute +w)(uv + vw + wu), 8 (uv + vw + wu)* > 34(uow)3, wu + +~+ > 3u) Third solution: In the same manner as in the Second solution, we reduce the problem to proving that (abe)? +2> 23 `ab— » Now, using Schur’s Inequality, we infer that 9ab 23 2ab—À `a?< Thưa
Trang 968 | Vasile Cirtoaje | Prove that IÍ 0 < z < < z and z+ + z = zz +2, then a) (1— zø)(1— z)(1— zz) 3 0; b) ay < 1a’y? < = Jays lay" < 55 Solution: a) We have (l—ay)(1—yz) =1l-—ay—yz+ay?2z =1—a2y—yz+y(@t+y+2—-2) =(y-1* >0 and similarly (1—øz)—zz) =(1—z)“>0, (1—zz)(1— zu) = (L— ø) >0
So the expressions 1 — xy, 1 — z and 1 — zz have the same sign
b) We rewrite the relation x+y+z = xyz+2 as (1—x)(1-y)+(1-z)(1—-zy) = 0 Ifz > 1 then z > # > z > 1 and so (I — z)(1 — ø) + (1— z)(1— zg) > 0, impossible So we have x < 1 Next we distinguish two cases 1) xy < 1; 2) zy > I1
32 1) ty <1 We have ay <a <1and ay’ <a<1< 5
2) xy > 1 From y > ,/ry we get y > 1 Next we rewrite the relation x + y+2z=
0z + 2 as # + T— 2 = (z0 — 1)z Because z > # gives # + — 2 > (z0 — l)U, (@T— 1)(2—#— z) > 0 so 2> z(1+ 0) Using the AM-GM Inequality, we have
1+ > 2Vÿ and 1+ = 1+ 548 5301-32 Thus we have 2 > 2e,/y and 2 32 2>3z = which means that xy < 1 and xy? < 7 32 The equality x?y = 1 takes place when x = y = 1 and the equality x?y”? = 2z 2
takes place when x = 3° y= r=2
Trang 10which is the same as 12(y+z) > 5+9y?+82?+ 1l2yz @ (2z—-1)?4+ (8y+2z—-2)? < 0,
which is clearly impossible Thus, the conclusion follows
70 | Gabriel Dospinescu, Marian Tetiva ] Let x,y,z > 0 such that
z + +Z—+xUz Prove that
(— 1)(w— 1)(z— 1) < 63 - 10
First solution:
Because of x < xyz > yz > 1 (and the similar relations xz > 1, zy > 1) at most
one of the three numbers can be less than 1 In any of these cases (2 < 1, y > 1, z > 1 or the similar ones) the inequality to prove is clear The only case we still have to analyse is that when x > 1, y >1 and z> 1
In this situation denote x-l=a,y-—1=6,z-l=c Then a,b,c are nonnegative real numbers and, because zœ=a+l,=b+l,z=c+], they satisfy a+1+b6+1+c+1=(a4+1)(b4+1) (c+), which means abe + ab+act+ be = 2 Now let « = Wabc; we have ab + ac + be > 3V abache = 32’, that’s why we get a + 34? <2 (x41) (x? 4+ 22-2) < 06
& (x +1) (ec +1+ V3) (ec +1- V3) <0
Trang 11Second solution:
Like in the first solution (and due to the symmetry) we may suppose that x > 1, y > 1; we can even assume that « > 1, y > 1 (for x = 1 the inequality is plain) Then we get xy > 1 and from the given condition we have +%+1 — 1` The relation to prove is (z — 1)(T— 1) — 1) < 6v3-— 10 âđ â 2#z (s + œz + yz) < 6V3 — 9,
or, with this expression of z,
day" — ay zy—-1 — (ex +y) —** <6V3-9 6 xy —1 & (xy —x —y)* + (6V3 — 10) zụ < 6/3 -— 9,
after some calculations
Now, we put x =a+1,y=0+41 and transform this into a2b? + (6v3 10) (a +b-+ab) —2ab>0 But
a+b> 2Vab and 6/3 — 10> 0, so it suffices to show that
ab? + (6V3 - 10) (2Vab + ab) — 2ab > 0
The substitution t = Vab > 0 reduces this inequality to tt 4 (sv3 — 12) # +2 (6v3-10)t> 0, “ f+ (6v3 - 12) t+2(6V3—10) > 0 The derivative of the function f(t) =0 + (6V8—12)++ 2 (6V8— 10), + >0 f#@=3 ( — (v3 — )) and has only one positive zero It is \/3 — 1 and it’s easy to see that this is a minimum is point for f in the interval [0, co) Consequently f(t) > f(V3-1) =0,
and we are done
A final observation: in fact we have
Trang 12which shows that ƒ (£) > 0 for £ > 0
71 | Marian Tetiva | Prove that for any positive real numbers a, b,c, a®—b bì— c ce —a*| _ (a—b)* + (b—0)* + (c— a)" — 4 * a+b b+c c+a Moldova TST, 2004 First solution: First of all, we observe that right hand side can be transformed into a3 — bŸ 3 x8 1 1 3 «3 1 1 2, a+b — (a -)(5-—) +0 -' (TT) 7 (a —b)(e— b)\(a—0) (> ab) (œ + b)(ø + c)(b + c)
and so we have to prove the inequality
|l(a — ð)(b — e)(ce— a)|(ab+ be+ ca) _ 1 2
(a+b)(b + e)(e+ 4) <5 (Ye _ 80)
(œ+b+ e)(ab + be + ca) and
It is also easy to prove that (ø + ð)(b + c)(e+ a)
so we are left with IV <O1 Go oa (X2(4— 9?) > [« -%| Using the AM-GM Inequality, we reduce this inequality to the following one (3) >|H«=9| This one is easy Just observe that we can assume that a > 6 > ¢ and in this case it becomes 3 (œ — b)(a — e)(b — c) < 2 and it follows from the AM-GM Inequality (a+b+ ø)Ÿ Second solution (by Marian Tetiva):
It is easy to see that the inequality is not only cyclic, but symmetric That is why we may assume that a > b > c > 0 The idea is to use the inequality
2 2
y winyty x
4+ 5 > ——— —— >†/+†+—
eros x+y <9 5
Trang 13That is why we can write a? — b? a’? + ab + b2 b + be -L c2 gŠ -L ae - c2 » a+b = (ape teeth a+b gether Và j2 Taote ` (a-b) (0+ $) +00) (c+ 5} — &=9 (ø+ 3) = 3692 4 In the same manner we can prove that ng < 5 ”(ø — b)? œ+b — 4 IV
and the conclusion follows
72 | Titu Andreescu | Let a,b,c be positive real numbers Prove that
(a° — a? +3)(0? —b7 4+3)(2 —c +3) > (at+b+e)®
USAMO, 2004
Solution:
We start with the inequality a° —a2+3>a?+2 6 (a?—1)(a?—1) > 0 Thus,
it remains to show that 3 II) > (2+) Using the AM-GM Inequality, one has a3 1 1 3a + + > 3 3 3 = ajt+2 bb) 4+2 42 #/[(a? + 2) We write two similar inequalities and then add up all these relations We will find that 3 [Ie +2> (Sa),
which is what we wanted
73 | Gabriel Dospinescu | Let n > 2 and #1,#s, ,#„ > 0 such that
DÓI ao
2 - = 2 4+——
(Sot): (a) owe mCn
Trang 14Solution:
In this problem, a combination between identities and the Cauchy-Schwarz Inequality is the way to proceed So, let us start with the expression Li 4; ° (= 4th 2) | 1<i<j<n t wi We can immediately see that > (r3) *j Li 1<i<g<n l| Z mm SI + SIS | |S | Hs S| + œ —_— l| 1<i<jcn
- (E82) Es) E3)~
Thus we could find from the inequality 2 ` (+3 -›) >0 — # 1<i<j<n (9) a)ae~ Unfortunately, this is not enough So, let us try to minimize 2 Ly %¿ > (r?) x 1<i<j<n J that This could be done using the Cauchy-Schwarz Inequality: (eens Lj %; n 2 3%; x; 2 c2, (Ee) Xj — Li 2; Because » (= a 2) = 1, we deduce that ‡<?<j<n j vi
Sa) (Ea) eras
which is what we wanted to prove Of course, we should prove that we cannot have equality But equality would imply that x, = v2 = - = &,, which contradicts the
É⁄)82)-e
Trang 1574 | Gabriel Dospinescu, Mircea Lascu, Marian Tetiva ] Prove that for any po- sitive real numbers a, Ö, c,
a2 + b2 + e? + 2abe+ 3 > (1+ ø)(1+b)(1+ e)
First solution:
Let ƒ(ø,b,e) = a? +? + c2 + abe+ 2— ø—b— e— ab— be — ca We have to prove
that all values of f are nonnegative If a,b,c > 3, then we clearly have - + : + : < 1, which means that ƒ(ø,b,e) > a? + b + cẰ+2—a—b—c >0 So, we may assume
b
that a < 3 and let m = — Easy computations show that f(a, b,c) — f(a,m,m) =
(3 — a)(b—c)* 1 > 0 and so it remains to prove that f(a,m,m) > 0, which is the same as (a+1)m? —-2(a4+1)m4+a?—a4+2>0 This is clearly true, because the discriminant of the quadratic equation is —4(@ + 1)(a — 1)2 <0 Second solution: Recall Turkevici’s Inequality z* +? + z2 +! + 2xuzt > z2? + 2z? + z2) + 12x? + z2 + 2£
for all positive real numbers x, y,z,t Taking t = 1,ø = #2,b = 9Ÿ,z = z and using the fact that 2/Vabc < abc + 1, we find the desired inequality
75 | Titu Andreescu, Zuming Feng | Let a,b,c be positive real numbers Prove
that
(2a+b+c)— (2b+a+c) (2c+a+b)? <8
2a2 + (b+c)? 26?+(at+c)? 2c3+(a+b)3 — —
USAMO, 2003 First solution:
Because the inequality is homogeneous, we can assume that a+ b+c= 3 Then
(2a+b+c)? _ a2 + 6a+9 _ M42 4a+3 ) <
Trang 16Second solution: b + b Denote x = re y= 2 eS We have to prove that a c (x + 2)? 2z +1 5 (2-1)? 1 WT <g6 <=e©ŠS `" ` ` an ——>- But, from the Cauchy-Schwarz Inequality, we have ye ` (z++z—3)? z?+2 —z?+?2+z?+6 It remains to prove that Q(x? + y? + 2? 4 Qay + 2z + 2z+ — 6 — 6y — 6z+9) >z?+?+z”+6© ©z?+ˆ°+z”+4(zụ + 0z + zz) — 12(œ + + z) + 123 0
Now # + z + zz > 3/+22z2 > 12 (because #z > 8), so we still have to prove that
(xtytz)?+24—12(2+y+z) +12 > 0, which is equivalent to (x +y+z-—6)? > 0,
clearly true
76 Prove that for any positive real numbers x,y and any positive integers m,n,
(n—1)(m—1) (a? ty™*") + (mtn—-l)(ary"+ary™) > mnlartr—ly tytn" 7), Austrian-Polish Competition, 1995
Solution:
We transform the inequality as follows:
mx —y)(a™ Fr" — y™*P—1) > (mtn —W(a™—y™)(a" —y") & .m+n—] _ ụ n1 gm — m coy nr n ứn +? — 1) — U) — m(œ— W)_ n(z — 0) (we have assumed that x > y) The last relation can also be written # x x ey) | min tars | ear f t” "dt ¥ ¥ Ụ
and this follows from Chebyshev’s Inequality for integrals
Trang 17with x,y, 2,t,u > 0 It is clear that 1 + 1 atabe — y tt ~ 1 tsi 1+ ab + abcd Ta T+ + Z U tạ 1 1 1 1 Writing the other relations as well, and denoting — = a,, — = aa, T— = đa, TT a4,-= x y Z u as, we have to prove that if a; > 0, then 1 ` _ 0a 1d „1 a, +agt+a5 - 3 Using the Cauchy-Schwarz Inequality, we minor the left-hand side with 4S? 2S% — (ag + a4)? — (a, + a4)? — (a3 + a5)? — (ag + a5)? — (a, + a3)?’ 5 where S = So ai By applying the Cauchy-Schwarz Inequality again for the
denominator Of the fraction, we obtain the conclusion
78 | Titu Andreescu | Prove that for any a,b,c, € (0, 3) the following inequality
holds
sỉn ø - sin(œ — Ö) - sin(œ —e) sinb-sin(6 — e) - sin(b — ø)_ sinc- sin(e — a) - sin(e — b) >0 sin(b + c) sin(c + a) sin(a + b) —
TST 2003, USA
Solution:
Let x = sina, y = sinb, z = sinc Then we have x,y,z > 0 It is easy to see that the following relations are true:
sina - sin(a — b)- sin(a — c) - sin(a + 6) « sin(a +c) = x(a? — y?) (a? — z”)
Using similar relations for the other terms, we have to prove that:
> 2(2? — y°)(a? —y*) > 0
With the substitution 2 = /u, y = Vv, z = Vw the inequality becomes » Vu(u — v)(u — w) > 0 But this follows from Schur’s Inequality
79 Prove that if a,b,c are positive real numbers then,
Va4 + b1 + c+ + Veh? + Pet 2a? > Va3b+ Bet Gat Vab? + be? + ca
KMO Summer Program Test, 2001
Solution:
It is clear that it suffices to prove the following inequalities
Trang 18and
(Soot) (Shas?) = (Sad) (2á)
The first one follows from Schur’s Inequality
So at+abeS a > S a?b+ áp
Sab? > abe S~ a
The second one is a simple consequence of the Cauchy-Schwarz Inequality:
and the fact that
(ab + Betcha) < (ab? + b?c? +? a7)(at + bt 4+ &*) (abŸ + beŠ + ca®)? < (a2bŠ + b°e? + e?a?)(a* + b + €)
80 | Gabriel Dospinescu, Mircea Lascu | Eor a given ø% > 2 find the smallest
constant k, with the property: if a1, ,@, > 0 have product 1, then a1 Q2 4 4203 4 4 Andy, <k (a? +a2)(az +a) (a3 +.a3)(a? + a2) (a2 + øi)(aˆ + an) ” Solution: 1 Let us take first aj = đứa = - =ữạ_ 1 —=#,0ạ = —- We infer that en bk > Qe2r-1 + m„ — 2 ` 7+ — 2 "= amETTTj@TT+T) ”(+z)” ”1+ø for all z > 0 Clearly, this implies k, > ?— 2 Let us prove that ø — 2 1s a good
constant and the problem will be solved
First, we will prove that (2? + y)(y? + ©) > zw(1 + z)(1 + g) Indeed, this is the same as (x + y)(x — y)” > 0 So, it suffices to prove that
1 1 1
+ fess + —— < n-2
(1+a1)(L+az) (1+ a2)(1 +43) (1+ an)(1+ ai) ~
Trang 19So, we have to prove the inequality n S- Chee >1, ha Mek + €p41) (p41 + Le+2) Using the Cauchy-Schwarz Inequality, we infer that n 2 de n 2 +1 k+1 > 2„ (tp + #g+1)(#g+q + #eg+2) — xẽ k=1 (tp + Le41)(Le+1 + Le42) k=1 and so it suffices to prove that n 2 Th nr Tr » «| > Sox; +23 7i + 0a k=1 k=1 k=1 k=1 But this one is equivalent to n n 2 › Liv; > 2 › %g#k+1 + ) 3g'k 3 1<i<j<n
and it is clear Thus, k, = n — 2
81 [ Vasile Cirtoaje | For any real numbers a, b,c, x,y, z prove that the inequality holds az + bụ + cz + (a3 + b2 + c2)(z2 + 2+ z3) > s(œ+b+e)(z++z) wl bo Kvant, 1989 Solution:
far ty? +22 v Hung
Let us denote # = \/ —————: Using the substitutions « = tp, y = tq and a3 -+ b2 + c2 Š Poy = tq z = tr, which imply C4+PtP=apPigtr The given inequality becomes os 2 aptbgter+a° +b +e > 5 at+b+c)(p+q+r), (at+p)?+(b+q)?4+(et+r) > =(atb+e\(ptqtr) Wl A Since 4(a+b+e)(p+qg+r)<[(a+b+e)+(p+aqg+r)}È
it suffices to prove that
(a+ p)? + (b+q)? +(ce+r) > =[(at+p) + (b+ q) + (ctr)
Wl]
—¬
Trang 2082 [ Vasile Cirtoaje | Prove that the sides a, b, c of a triangle satisfy the inequality
3(0 4242-1) 50(2 4622) b ¢ oa a b oe
First solution:
at c After
We may assume that c is the smallest among a, b,c Then let x = b— some computations, the inequality becomes
(3a—2e)z°+ (ø +e- 1) (a—c)? > 0 © (3ø—2e)(2b—a—e)?+(4b+2e—3a)(a—e)Ÿ > 0 which follows immediately from 3ø > 2c, 4b + 2e — 3ø = 3(b+e— ø)+b—ec>0
Second solution:
Make the classical substitution a =y+2,b=z++2,c=2x+y and clear denomi- nators The problem reduces to proving that
#2 + 9 + z3 + 2(z?u + 0 )z + z +) > 3(z2 + uz + zz”)
We can of course assume that x is the smallest among x,y,z Then we can write y=xt+m,z=2+Nn with nonnegatives m and n A short computation shows that
the inequality reduces to 2x(m? — mn +n?) + m3 +n? + 2m?n — 3n?m > 0 All we need to prove is that m? +n? + 2m?n > 3n?m | (n—m)> —(n—m)m? +m? > 0 and this follows immediately from the inequality t? + 1 > 3, true for t > —1
Trang 21But this one is not very hard, because it follows immediately by multiplying the 1 1 142) > 1+ n—-1 —— II ( #7 JIL, obtained from Huygens Inequality inequalities n—-1 Second solution: We will prove even more, that - 1 "?—1\” xy 1 l(-3)>(% 'lrz x ¿=1 , It is clear that this inequality is stronger than the initial one First, let us prove that
This follows from Jensen’s Inequality for the convex function f(x) = In(1+ 2) — In(1 — x) So, it suffices to prove that 1) - n”—1\" —>——:||a-z”>|——] - ñ lƯ PT) But a quick look shows that this is exactly the inequality proved in the solution of the problem 121
84 | Vasile Cirtoaje, Gheorghe Eckstein | Consider positive real numbers #1,Z2, ,„ such that #+zs #„ = l Prove that 1 1 1 + + +®————<Il nm—-lt+2a n—-14+ 22 m=— Ì + #„ TST 1999, Romania First solution:
Suppose the inequality is false for a certain system of n numbers Then we can find a number & > 1 and n numbers which add up to 1, let them be a;, such that
1
——- = ka; Then we have n—-1+2;
Trang 221
We have used here the fact that a; < n1 Now, we write | — (n — l)ay = by and
nN —
we find that ` b„ = 1 and also I[a — by) < (n — 1)”bị b„ But this contradicts
the fact that for each 7 we have 1—6; = 6) +-+-+bj-1 + bj41 +-:- +b, > (n-1) "YX by .b3-10;41 On Second solution: Let us write the inequality in the form #1 33 Ln - + +————> nm-lt+a, n-1+%2%2 nm—-14+2p This inequality follows by summing the following inequalities 1—+ 1—+ — — 1—+ 1—+ 1—+)``` — — 1—+ — 1
nh T1 cự hượ tự, dan ” n— Tần ph cự ta, 4+n The first from these inequalities is equivalent to Lt 1—+ 1—+ — + a, "+a, "+ +4n "> (n-1)x," and follows from the AM-GM Inequality Remark
Replacing the numbers 21, %2, ,%, with i Ty respectively, the in-
equality becomes as follows ma "
1 1 1
1+(m-— 1)zi T 1+(n—1)za T::*TT@n-1zaÊ \
85 [ Titu Andreescu |] Prove that for any nonnegative real numbers a, b,c such
that a? + 6? + c? + abc = 4 we have 0 < ab+ be + ca — abc < 2
USAMO, 2001
First solution (by Richard Stong):
The lower bound is not difficult Indeed, we have ab + bc + ca > 3Va2b2c2 and thus it is enough to prove that abc < 3Va2b2c2, which follows from the fact that abc < 4 The upper bound instead is hard Let us first observe that there are two numbers among the three ones, which are both greater or equal than 1 or smaller than or equal to 1 Let them be b and c Then we have
4 > 2be +a” + abe > (2—a)(2 +a) > bc(2 +a) > be < 2-4
Thus, ab+ bc+ca-—abe < ab+2—a+ac—abc and it is enough to prove the inequality
qb + 2— a+ ac— be < 2© b+c— be< 1© (b— 1)(c— 1) >0, which is true due to
Trang 23Second solution:
We won’t prove again the lower part, since this is an easy problem Let us con- centrate on the upper bound Let a > 6b > cand leta=ax+y,b=a2-—y The
hypothesis becomes x?(2 + c) + y?(2 —c) = 4—c? and we have to prove that (2? — y?)(1—c) < 2(1 — 2c) Since y? = 2+ e— — the problem asks to
4 2 4— 2 4
prove the inequality dot 4) —c) < 2(1 — cx) Of course, we have c < 1 e
and 0 < y? = 24+c- 37? z? 2-c > x < V2-c (we have used
the fact that a > 6 > y > O and b 0 > x« > y > 0) Now, consider the
4z2 — (4— c?)
function ƒ : [0,v2— e| > R, f(x) = 201 - cx) - — ze —c) We have
l-e
f'(x) = -2c— 82- 5 < 0 and thus f is decreasing and f(x) > f(W2—c) So, we
have to prove that ƒ(v2—c) > 06 2(1-evV2 — c) > (2-0)(1-c) 6 3 > c4.2V2 —6 6 (1-V2—)? > 0 clearly true Thus, the problem is solved 86 | Titu Andreescu | Prove that for any positive real numbers a, b, c the following inequality holds at+tb+e 3 ~ Vabe < maz{(VJa — vb)", (Vb — Ve)”, (We — Va)’} TST 2000, USA Solution: A natural idea would be to assume the contrary, which means that a+0+€_ #1 <œ+b- 2Vab 3 b “TT —— Ÿabe<b+e~ 9Vữc a+b+€c 3 — Vabe < e+ø— 2V(a
Adding these inequalities, we find that
a+b+e—3WVabe > 2(a+b+c¢—Vab— Vbe— Vea)
Now, we will prove that ø+b+ e— 3Ä⁄abe < 2(a+b+e— Wab — Vbe — ca) and the
problem will be solved Since the above inequality is homogeneous, we may assume
that abc = 1 Then, it becomes 2Vab + 2V/bc + 2\/ca — a — b—c < 3 Now, using
Trang 2487 | Kiran Kedlaya | Let a,b,c be positive real numbers Prove that
a+ Vab+ Vabe _ af a+b a+b+e
3 — 2 3 ‘
Solution (by Anh Cuong):
We have that
a+Vab+ Vabe < a+ §/ar° + Yabe Now we will prove that | b / b b at lao + Vabe < Va os By the AM-GM Inequality, we have Le 2a + 3a 3/1 2a 8a < a+b a+b+ce a+b a+b+ec_ 3 , 3b 2 — 31.1 _ 3U « Ta b+c atb+te7 3 , Le 2b + 3¢ sy, 26 3c a+b_ a+b+ec a+b a+b+ec— 3 l Now, just add them up and we have the desired inequality The equality occurs when a=b=c 88 Find the greatest constant & such that for any positive integer n which is not a square, |(1 + /n)sin(r/n)| > k Vietnamese IMO Training Camp, 1995 Solution:
We will prove that is the best constant We must clearly have k <
Trang 251 i) The first case is when {,/n} < 3 Of course, 1 {vn} 2 Vin Vn— T= —" 3 x and because sin x > x — S we find that sin(z{Wn}) > sin ———— > (=) — 1 (——): —— Wn—l+vn — \vn— L1+ vn 6 \VøT— 1+ vn 7T Let us prove that the last quantity 1s at least ———— This comes down to P 4 y 2(1 + Jn) 2+ vn=vn=T x2 1+ vn 3(Vn+wn=T)”” or 6(⁄n + Vn — 1)? + 3(VWn + vn — 1) > a?(1 + Vn) and ït is clear 1 1 ii) The second case is when {./n} > 3 Let er = 1— {Wn} < 5 and let n = 1 k? + p,1 < p < 2k Because {,/n} > 5 = p > k+l Then ït 1s easy to see that 1 x > —————— _ k+l+vk?+2k and so it suffices to prove that TT TT sin > k+14+Vk?+2k ~ 2114 Vk? +k) 3 Using again the inequality sing > x — 6 we infer that, 2Vk2+k— Wk2+2k— k+ 1 5 > 1” 5: l+Vk° +k 3(1+k+ Vk? + 2k) But from the Cauchy-Schwarz Inequality we have 2Vk? + k— Vk? +2k—k>0 2 Because the inequality 1 +k+vk?+ 2E)” > > (1 + Vk? + i) holds, this case is also solved 89 [ Dung Tran Nam ] Let 2, y,z > 0 such that (27 + y+ z)° = 32zryz Find the ety? +24 minimum and maximum of 7: (ct+y+z) Vietnam, 2004
First solution (by Tran Nam Dung):
Trang 26at +y* +24 44 ¬ = (a? ty? +2) 2Š” z?y2 — (16— 2 So ay)? — 23 #)Ÿ + 4zz(œ + + z) = = 2a?— 64a + 288, extremal values of Now, we have 2
where a = xy+yz+za Because y+z = 4—2 and yz = —, we must have (4— #) x 2> which implies that 3— /5 < « < 2 Due to symmetry, we have 2, y,z € [3 — V5, This means that (a — 2)(y — 2)(z — 2) < 0 and also (— 3+ V5)( - 3+ V5)(z — 3+ v5) > 0 ? 93 |%œ Clearing paranthesis, we deduce that ae „#21 z2 +t+z? — (a— 16)?— 112 44 7 128 —1 9 1 1+ v5 383 — 165V5 (3- vs + vỗ 4) But because , we find that the extremal values of the expression are 256 » =, attained for the triples 128 yD respectively (2,1, 1) Second solution: As in the above solution, we must find the extremal values of x? + y? + z? when 1
#-+ + z = l,zz = 39° because after that the extremal values of the expression xe’ + y* + z* can be immediately found Let us make the substitution x = T1 =
b 2 b2 4 2
T2” 7 where abe = 1,a +64 2c =4 Then 2? + y?4 27 = ———— and so we
2
must find the extremal values of a? + b? + 4c” Now, a? +b? + 4c? = (4— 2e)3— ~ +4e? c
and the problem reduces to finding the maximum and minimum of 4c? — 8c — 1 where there are positive numbers a,b,c such that abc = 1,a +64 2c = 4 Of course, this 3— võ 2 , 1} But this reduces to the study vỗ a 1 — of the function f(z) = 4x? — 8z — — defined for x 4 4 comes down to (4 — 2c)? > -, or toc € | c
, which is an easy task
90 | George Tsintifas ] Prove that for any a,b,c,d > 0,
(a + b)?(b + e)?(e+ đ)?(d+ a)) > 16a?b?c2d?(œ + b+ e+ đì!
Trang 27Solution:
Let us apply Mac-Laurin Inequality for
x = abc, y = bed, z = cda,t = dab We will find that
S| abe ° ` 3 abc - bed - của _ a2b2c2đ2 ` a 7
4 — 4 7
Thus, it is enough to prove the stronger inequality
(a+ b)(b+c)(e+d)(d+a) > (a+b6+c+4d)(abe + bed + cda + dab)
Now, let us observe that
(a + b)(b + e)(e + đ)(đ + a) = (ae + bđd + ad + be) (ac + bd + ab+ cd) = = (ac + bđ)? + 3 a?(be + bd + cả) > 4abcd + È) a®(bc + bd + cd) = = (a+b+c+d)(abe + bed + cda + dab)
And so the problem is solved
91 | Titu Andreescu, Gabriel Dospinescu |] Find the maximum value of the ex-
pression
(ab)” , (bc)" | (ca)”
l—ab 1-—be 1 -ca
where a,b,c are nonnegative real numbers which add up to 1 and n is some positive integer Solution: First, we will treat the case n > 1 We will prove that the maximum value is 1 3 gna" It is clear that ab, bc, ca < 1 and so (ab)” + (bc)” (ca)” _ 4 1—ab 1—be l—eca < 3 ((ab)* + (be)* + (ca)*) 1
Thus, we have to prove that (ab)” + (bc)” + (ca)* < 7 Let a the maximum among
a,b,c Then we have 1 m >a"(1—a)" =a"(b+c)" > a"b" + a"e" +naPb"—1e > a"b" + phe" + cha", 1 So, we have proved that in this case the maximum is at most 3.an=1' But for 1
a=b= 5 c = 0 this value is attained and this shows that the maximum value is for n > 1 Now, suppose that n = 1 In this case we have
ab 1
`
3.4n=1
Trang 28Using the fact that a+b+c= 1, it is not difficult to prove that 1 1 1 3-2 » ab + abe 1—ab ẹ T—be* 1-ca | 1— Sab + abe - abc? b 1 ˆ ob < a With the above observations, this reduces to ` ab < 3 — 27(abc)? + 19abc 11 But using Schur’s Inequality we infer that 1 3 ab < oo
We will prove that »
and so it is enough to show that
9abc + 1 < 3 — 27(abe)? + 19abec
4 — 11
& 108(abc)* + 23abe < 1,
1
which is true because abc < a7
Hence for n = 1, the maximum value is 3 attained for a = b= ce 92 Let a,b,c be positive real numbers Prove that 1 + 1 + 1 > 3 a(l+b6) b+e) c(1+a) — Ÿabe(1+ Vabe) Solution: The following observation is crucial 1 1 1 l+abe+a+ab (+ abe) (T+ ota) +3 = 2 annp _ l+a b(c + 1) 7 aa td I+b ` We use now twice the AM-GM Inequality to find that ys +® e1) > 3— + 3Ÿabe a (1+) 1+b — VYabe And so we are left with the inequality 3 + 3Vabec — 3 3 Vabc 3 > 3
1+ abe ~ Vabc(1 + Vabc)
which is in fact an identity!
93 [ Dung Tran Nam ] Prove that for any real numbers a, b,c such that a? +6? +
cẰ =9,
2(z+b+c)— abc < 10
Trang 29First solution (by Gheorghe Eckstein):
Because max{a, b,c} < 3 and |abc| < 10, it is enough to consider only the cases
when a,b,c > 0 or exactly of the three numbers is negative First, we will suppose that a,b,c are nonnegative If abc > 1, then we are done, because
2(a+b+c) — abe < 24/3(a2 + b2 + c2) — 1 < 10
Otherwise, we may assume that a < 1 In this case we have
b2 -+ œ2
St +) =dác <9 [s3 = = 2a + 2V18 — 2a? < 10
Now, assume that not all three numbers are nonnegative and let c < 0
Thus, the problem reduces to proving that for any nonnegative x,y,z whose sum of squares is 9 we have 4(a + y — z) + 2ayz < 20 But we can write this as
(z—2)”+(/—2)”+(z—1)Ÿ > 2zwz — 6z — 2 Because 2zz—6z—2 < z(y?+z”?)—6z—2 =
—Z + 3z — 2 = -(z — 1)?(z + 2) < 0, the inequality follows
Second solution:
Of course, we have |a|, |b|,|c| < 3 and |a + b + c|,|abe| < 33 Also, we may
assume of course that a,b,c are non-zero and that a < ö < e le < 0 then we have
2(a+b+e)—abe < —abe < 3V3 < 10 Also, if a < 6 < 0 < cthen we have 2(a+b+c) <
2c <6 < 10+ abc because abc > 0 If a << 0 < b < c, using the Cauchy-Schwarz
Inequality we find that 26+ 2c—a < 9 Thus, 2(a+6+c) = 2b+2c—a+3a <9+3a
and it remains to prove that, 3a — 1 < abe But a < 0 and 2be < 9 — a”, so that it
— > 3a— 1© (a + 1) (a — 2) < 0, which follows So, we
Just have to threat the case 0 < a < b < c In this case we have 2b + 2c + a < 9 and
2(z+b+c) <9+a So, we need to prove that a < 1 + abe This is clear iÍ a << 1 and
if a > 1 we have b,c > 1 and the inequality is again Thus, the problem is solved
remains to show that
Trang 30we get
0z + ø + z + zz >3 4,
> 0 we distinguish
(c= 1)*
with equality if and only if abe = 1 Because y+z = tL bt
two cases a) x> 0, yz < 0; b)x> 0,y> 0,z> 0 ,
a) x> 0, yz < 0 We have xyz < 0 and Írom #2 + ø + z + zz > 4 we get z Ặ+z+zz> 4> 3 b) x> 0,y> 0,z> 0 We denote z + yz + zx = 3d? with d > 0 From the mean inequalities we have z + z + zz > 33/+z292z2, from which we deduce that xyz < d? On the basis of this result, from the inequality 2z + #U + 0z + zz > 4, we obtain đ + 3d? > 4, (d— 1)(d+ 2)? > 0,d > 10 z + z + zz > 3 With this the given inequality is proved We have equality in the caseea=b=c=l Second solution: Letu=a2+lv=y+1,w=2z+4+1 Then we have 1 uw =ututw+tabe+ Tà >utvtw+2 abe
Now, consider the function f(t) = 2¢°+t?(u+v+w)—uvw Because Jima f(t) = œ and
f() < 0, we can find a real number r > 1 such that f(r) = 0 Consider the numbers
m=-,n= 0 —,p= — They verify mnp = m+n+p+2 we deduce from problem 49
that Am +np-+pm È > 2(m-+n.+p) > ưu + 0u + tu > 2r{(u + +) > 2(u +0)
But because u = x+1,v = y+1,w = z+1, this last relation is equivalent to sy +yz+ zx > 3, which is what we wanted
95 [ Gabriel Dospinescu |] Let n be an integer greater than 2 Find the greatest