Tuyển tập đề thi vô địch bất đẳng thức thế giới P3

So, we may assume that a,b,c are the side lenghts of a triangle ABC.. With the usual notations in a triangle, the inequality becomes 2 a?. Of course, this is true for any other variable,

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)?

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)? >

(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:

Clearly, if one of the factors in the left-hand side is negative, we are done So,

we may assume that a,b,c are the side lenghts of a triangle ABC With the usual notations in a triangle, the inequality becomes

2 (a? +b? +c’) < abc(ab+be+ca) & (at+b+c)(ab+bc+ca)R? > abc(a* +b? +c’)

atb+e

But this follows from the fact that (a + 6+ c)(ab + bc + ca) > 9abc and

0< OH? =9R?-— a2 — b — cẺ

58 | 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

abe Ya+ — +) a c+ 5 > 2(Soa+ am)

But this follows from the inequalities

œ“bc + P > 2ab, “ca + — > 2bc,c“ab + ọ > 2ca

a and

Of 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+ 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

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

> 1 (of course, we may

and so it is enough to prove that the last quantity is at least 1 But it follows from

Using the substitution 6 = a+1(6> 2) andz= -, y= Đ z= - (abc = 1) the

inequality becomes as follows

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 =

(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

where we used the inequalities

(Siz) >3 (So zy) and xyz < wa

66 | 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

i) If a,b,c are at least 1, then TT (“ 3 +1) >14505 5 » 2)

ii) If two of the three numbers are at least 1, let them be a and 6, then we have

Il a-1., peal Pal +2

Second 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

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)

68 | 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

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>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

69 | Titu Andreescu | Let a, b,c be positive real numbers such that a+b+c > abc Prove that at least two of the inequalities

2 eg 78 8g 28 Log a Ob Cc b ¢c¢ 4a ec a ob are true

TST 2001, USA

Solution:

; = Y, : = z Thus, we have #,,z > Ö and z# + z + zz > 1 and we have to prove that at least two of the inequalities 2z+3-+6z > 6, 2u+3z-+6z > 6, 2z+3z~+6y > 6 are true Suppose this is not the case Then we may assume that 2z-+ 3„-+6z < 6 and 2z+3z-+ổy < 6 Adding,

5-5

ễ, Thus, 12 > sẽ +9 +8z

1 The most natural idea is to male the substitution -— = x,

a

1

we find that 5a+9y+8z < 12 But we have x > +

Ù

which 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 + 34? <2 (x41) (x? 4+ 22-2) < 06

& (x +1) (ec +1+ V3) (ec +1- V3) <0

For x > 0, this yields

Second 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 ©®

© 2#z — (sự + 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

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

f(t) = (1 v3+1) (t+2v3- 2).

which 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)"

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

(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

act

That is why we can write

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,

which is what we wanted

73 | Gabriel Dospinescu | Let n > 2 and #1,#s, ,#„ > 0 such that

DÓI ao

(Sot): (a) owe mCn

Prove that

- (E82) Es) E3)~

Thus we could find from the inequality

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

assumption

74 | 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

(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 ) <

2a2+(b+c)? 3a2—-6a+9 3 2+(a—1)?/ —

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") &

and this follows from Chebyshev’s Inequality for integrals

77 Let a,b,c,d,e be positive real numbers such that abcde = 1 Prove that a+ abc + b + bcd + c+ cde + d + dea + e+eab

l+øaồ+abcd 1+bc+bcdc 1+ cd+cdea TI+de+decab_ ` 1+ ca + cabc

with x,y, 2,t,u > 0 It is clear that

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) —

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

So at +S oa? b? > S > ab + áp?

and

(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

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

+ fess + —— < n-2

(1+a1)(L+az) (1+ a2)(1 +43) (1+ an)(1+ ai) ~

Now, we take a, = „HỘ c.«a đựy — “ and the above inequality becomes