Inequality in international tests in mathematics

| Gabriel Dospinescu | Prove that for any positive reals a,b,c we have 10... Let x,y,z be positive real numbers satisfying the condition... lran, 1999 Solution: A quick look shows that a

Inequality in international tegts

in mathematics

————==-—=# =

b+c cta ath ?

J@ aod? + (Pano? + VE nna > 3?

2

holds for arbitrary real numbers a, Ö, e

Komal Solution 1:

Applying Minkowsky’s Inequality to the left hand side we have

Ja? + (1—b)?4+/e? + (1-6)? +e +(1—a)? > VW(3—a—b—e)3+(a+b+)

Denoting a+b+c=<2 we get

3\" _ 9

(3—a—b—e)?”+(a+b+~e)? =2(- 5) > 5 and the conclusion follows

Solution 2:

We have the obvious inequalities

Ja? + (1— ð)3 + W2 + (1— e)? + Ve + (1-a)? >

„ Jal+ll=ð[J+lt=el„ Jel+lt=a|

3Vv2 and the last quantity is clearly at least >

Observe that «? < x3 for x € (0,1) Thus we have

and

Summing up, we obtain

Vabe + {(1— a)(1— b)(1— e) < Vb-VWe+ VW1—b- V1—e< 1,

the last one being an obvious application of Cauchy-Schwarz Inequality

Solution 3:

Let a = sin?z,b = sin? y,c = sin? z, where z,y,z € (0, 3): The inequality becomes

sinz:siny:sinz+cosx:-cosy:cosz < Ì and it follows from the obvious inequalities

sinz-siny:sinz+cosx-cosy-cosz < sinx-siny +cosz-cosy = cos(x —y) <1

abc(a + 6)(b+ c)(e+ a) ~ 23-(a+b+c)®’

(Aenea)! — (@tbaOrdreray > (a+b)(b + e)(e + a)

as needed

A,

5 Let 21, 2%2, ,%, be the numbers from the interval (0,

mls ) such that tan x, +

1 Jon

Bosnia and Herzegovina, 2002 tan #a + + tanz„ < n= Prove that sin x, - sing sinz, <

Otherwise, the same inequality gives

and the conclusion follows

8 | Mircea Lascu | Let a,b,c be positive real numbers such that abc = 1 Prove

that

Tat et Fe 2 Vat vot ver 3

Gazeta Matematica

Solution:

From the AM-GM Inequality, we have

te ote 2Š: + [2+ -/)-

9 | Gabriel Dospinescu | Prove that for any positive reals a,b,c we have

10 [ Claudiu Coanda | Prove that for any positive numbers a, b,c we have

2+(at+bt+c) ~ 3%(at+bt+e?~ Ÿ⁄gở⁄2?t2e e/ a2b?2 ¢2

We combine these two inequalities and the problem is solved

and

11 | Hojoo Lee | Let a,b,c > 0 Prove that

Va4 + a?b2 + b + J+ PR +4 Vt+ a2 +a! >

> av 2a? + be + by 2b? + ca + eV 2c? + ab

Gazeta Matematica

Solution:

We start from (a? — 6”)? > 0 and we expand it and we rewrite it as 4a* + 4a7b? +

Ab* > 3a* + 6a?b2 + 3b† and we get a1 + a2b2 + b2 > Ba? + b)

Using this observation, we find that

(> a* + a?b2 + pi) >3 (Se)

But using Cauchy-Schwarz Inequality we get

(> avd Fhe) < (> a*) (> (2a? + be) ) <3 (> a)

and so the inequality is proved

12 | Darij Grinberg | If a, 6, c are three positive reals, then

and the inequality (1) is rewritten as 7 + B + GC +4 7A + B + =) > 15 Because

C A B 3/4 BC

A,B ; " 0, we have from AM-GM M-GM Inequality that atRt c> 3 “—+—+—>3Ÿ/<-—-= BC A

3 and 7 + 5 +— > 3, from where we get the desired inequality

An alternative solution for (1) would be using Cauchy-Schwarz Inequality, so

The equality occurs when ø = #, ö= z,e= and 2z > + z

15 [D Olteanu | Let a,b,e€ ¿ Prove that

b+e 2a+b+c c+a 2+e+a atb 2c+a+b— 4

16 [| Adrian Zahariuc | Prove that for any a,b,c € (1,2) the following inequality holds

Rp > sj ta- bral +b! > ab(a + b) © (œ — b)*(a + b) >0,

which is clearly true Writing the analoguous inequalities, and adding them up, gives

yy [Stet tert oF tee `

a;

and the problem 1s solved

20 The real numbers a1, @2, ,@100 satisfy the relation af + a3 + -+-+ ato g +

(ai + aa + - + øiog)° = 101 Prove that |ay| < 10, for all k = 1,2, , 100

Junior TST 2004, Romania Solution:

We will prove that |a,;| < 10 Let A = ag+a3+ -+aio99 Using Cauchy-Schwarz

A Inequality, we find that 101 > a? + 99 +(A + ai)” But it is immediate to prove

5(a7 + bŸ + c2) < 18abe + 6(a2 + bŠ + e?) — 6(ab + be + ca) + 1 ©®

© 1§abe + 1 — 2(ab + be + ca) + 1> 6(ab + be + ca) © 8(ab + bé + ca) < 24 1Sabc ©>

© 4(ab + be + ca) < 1+ 9abe > (1 — 2a)(1 — 26)(1 — 2c) < abe ©

& (b+c—a)(c+a—b)(a+b-—c) < abc, which is equivalent to Schur’s Inequality

22 Consider the positive real numbers x, y, z so that

a), b), c) Using well-known inequalities, we first have

ays = 2? +y? +2? > 38/22y222 = (xyz)? > 27 (xyz),

which yields xyz > 27; then

cy +02 +yz > 34) (xyz)? > 3V272 = 27 and

and, after a small calculation,

@Š + bŠ+c?+a+b+c+2= abe + ab + ac + be

If we put g = ab+ac+ bc, we have according to the same well-known inequalities

qg<a°+?+c°, V3qg<a+b+c and

Because (x + y + 2)? > 3(ay + yz + zx) we get 1+

so it suffices to prove that

r+yt+2Z

Mircea Lascu and Vasile Cartoaje

24 If the equation x* + ax? + 2x? + be + 1 = 0 has at least one real root, then a? +b? > 8

Tournament of the Towns, 1993

a? +b? >

25 | Manlio Marangelli | Prove that for any positive numbers x, y, z we have

3(z2 + y?z + 272) (ay? + yz? + za") > xuz(œ + + z)

3 2z+z2z+z?u z2+zz? + +? — 8/3 (y2z+ 22a + +?) - (wz2 + zx? + xy?)

and two other similar relations

It is immediate to prove that

| sin(2 + y)| < min{|cosz| + | cos y|,| sin z| + | sin y|}

and

| cos(x + y)| < min{| sin] + | cos y|,|sin y| + | cos z|}

Thus, we infer that

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 + 44 + 25)| < | cosx3| + | cos x4| + | cos x5|

and the conclusion follows.

27 | Ioan Tomescu | Let x,y,z > 0 Prove that

LYZ (1 + 3x) (x + 8y)(y + 9z)(z + 6) When do we have equality?

1

Š

Gazeta Matematica Solution:

First, we write the inequality in the following form

therefore 2ý — l <0 ©< 5 this meaning that ryz < 3°

b) Denote also s = x + y + z; the following inequalities are well-known

(cx ty +z) >3(ay +224 yz) and

(x ty tz)? > 27xyz:

then we have 2s? > 54xyz = 27 — 27 (ay + xz + yz) > 27 — 9s”, 1 e

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

2 > 3q¢=3(1-2p)>3(1-=) =-;

if we put g=aytauzt+ yz, p= xyz

Now, one can see the following is also true

3

q= + #Z + W2 3 1:

1 c) The three numbers #,,z cannot be all less than 5° because, in this case we get the contradiction

3 1

% + zz + z + 2xz < 1128 = 1;

because of simmetry we may assume then that z > 5

We have 1 = (224+ 1)ay+z(a@+y) > (2241) ay + 2z,/ey, which can also

be written in the form ((2z +1) /zy—1) (,/zy+1) < 0; and this one yields the inequality

1

4 < ———s

(2z + 1)”

2 W©e also have l = (2z + I)z + z(z + 0) < (2241) +) + z(Z + 9), conse- quently ((2z + 1) (a+ y) — 2) (@ + y+ 2) > 0, which shows us that

z+0> 2z+1 The inequality to be proved

1 1 1

—=+—=+—->4(z++z)

x sy Z can be also rearranged as

1 4z?— 1 —1)(2 œ+u)(=—4)> Z _ (22 )@z+)

a,b,c such that # = ———., = ——.z = And now a),b) and c) reduce

immediately to well-known inequalities! Try to prove using this substitution d)

Let #„„ = min{Z1,#a, ,#„} and #a;¿ = max{z41,#a, ,#„} ŠSuppose without loss

of generality that m < M Let

St = (@m — @m4i)? 4++++ (e@m—1 — eu)?

and

So = (t@u — @Ma4i)° + + (tp - @1)° + (a1 — z2)” + -*- + (#m—1 — Lm)*

k i/& 2 The inequality Soa; > = (> «| (which follows from Cauchy-Schwarz In-

Pa z1 w=1

equality) implies that

and the problem is solved

31 If a,b,c are positive real numbers such that abc = 2, then

(a2 + b2 + e?)? < (a+b+ e)(œŠ + b + c”).(2)

These two inequalities combined yield

(a? +b? +c?)(at+b+c)

a”-+b”+c7 >

6

` (aVb+c+ ware + cvø + Ð)? (3)

Using AM-GM Inequality we obtain

aVb+e+bWa+e+ecVa+b > 34 abe ( (a+b)( + e)(e+ a))

> 3V abcVRabe = 3Ÿ = 6.

Thus

(aVb+c+b/ate+evatb) > 6(avb+c+bVate+cVa+t b).(4) The desired inequality follows from (3) and (4)

Solution 2:

We clearly have

aVb tetbVetateVvatb< ⁄2(42 + b2 + e2)(a + b+ e)

Using Chebyshev Inequality, we infer that

J2at+P+e)(atb+e) < /6(a? +83 + 3) and so it is enough to prove that a? + 6° + c? > 3abe, which is true by the AM-GM Inequality We have equality when a = b =c= V2

32 Let x,y,z be positive reals with sum 3 Prove that

hence

ety? +2 +2(vet Vat V2) > 3(w + + 2) > 9

33 Let x,y,z be positive real numbers satisfying the condition

and the conclusion is plain

c) These inequalities are simple consequences of a) and b)

that x,y < 3° Or #,1 > 5 and we then have

1 (2z — 1)(2u—1)>0©z+— 2zu< 5

On the other hand, we can get

1 = a 4y? +274 2Qaeyz > Qayt 274 2cyz >

r= sin 5: y =sin =; z= sin 5 according to this, new (trigonometric) proofs can be given for such inequalities

and it is clear, because n > 3 and a; + aj41 + aj42 < a1 + ao +-+-+ a, for all 2

36 Let a1, a2, ,@n > 1 positive integers Prove that at least one of the numbers

“d2, °%/03, , “n=V@n, °/@1 1s less than or equal to V3

Adapted after a well-known problem Solution: ,

Suppose we have đc) > 33 for all i 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

n+l

#¿+1

Thus, using this observation, we find that THẾ, > 33 > d1 => aiy1 > a; for alli,

which means that a, < ag <-+- < Gn_1 < Gy < a1, contradiction

37 Suppose that a, < ag < < @» are real numbers for some integer n > 2 Prove that

ayas + A204 + 4+ a„a1 > aaa1 + a3as + + ad}

lran, 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?) > ca(c? — a’) & (c? — b°)(ac — be) < (b® — a®)(ab — ac) Because a < 6 < ¢, the last inequality reduces to ø(bŸ + ab + a2) < c(c? + be + b)

And this last inequality is equivalent to (ec — a)(a? +b? +c? + ab+be+ca) > 0 which

is clear

38 Find the maximal value of the expression x? +y?+2?—3zyz when x?+y?+2? =

= 1 and z,y, z are reals

we clearly have (1+2#)(#— 1)? < 1œ f?(3—2#) > 0 and thus |z3+”+z3—3zz| < 1

We have equality for # = 1, = z = 0 and thus the maximal value is 1

39 | Viorel Vajaitu and Alexandru Zaharescu | Let a, b, c be positive real numbers Prove that

ab + be + ca < Ì(a+b+e) s+b+2c b+ec+2a c+a+92b— 4 ‘

is true for all positive reals a,b and c

< qiat otc)

AO | Walther Janous | Let x,y and z be positive real numbers Prove that

= 3 ễ 1

r+ V10) 12) J+VỤ+2025) 2+VGtBGTy =

Crux Mathematicorum

Clearly, we may assume that a = min{a, b,c} and let us writeb =a+2,c=a-+y,

where x, y € [0, 1] It is easy to see that ả +6? + c? —3abe = 3ă#?T— zử)+z3 +

and ảb + b?c + c?a — 3abc = ăx? — ry + ỷ) + xỵ So, the inequality becomes 1+zỶ + > 3z?2 But this follows from the fact that 1 + z3 + 3 > 3z > 3z?9, because 0 < #,y < 1

43 Prove that if a1, @o, ,@n,01, ,b, are real numbers between 1001 and 2002

such that ả + a3 + -+ a2 = 67+ 03 + -+b?, then we have the inequality

and also

oO and the inequality — -— i> 1 +5 — allow us to

Now, the observation that B = a, Dy 2 ; ỉ

Using (1) and (2) we find that

which is the desired inequality

44, Let a,b,c > 0 with a+6+c= 1 Show that

a2 +b +e b+e c+a

C+a

> 2

a+b — Solution:

Using Cauchy-Schwarz Inequality, we find that

The last inequality can be transformed as follows

14 (Soe) > 2S P4042) ab & 14+ (Soa) >

2S oa -25°a +25 ab = (Sve?) +25 a? > Soa’, and it is true, because

\2 » a?

rey de 3 [ Calin Popa |] Let a, b,c be positive reals such that a +b+c¢= 1 Prove that

Jb / Lab With the substitution « = Tư = se — ,/© the condition a +b+c=1

where we used the obvious inequalities

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

46 Let a,b,c > 0 such that at + b4 + ct < 2(a”bŠ + b”c? + 2a”) Prove that

a® <ab+tac, b? <be+ba, 2 <cat+cb and the conclusion follows

47 Prove that for any a,b,c, x,y,z positive numbers with x + y+ z= 1 we have the inequality a2 + bụ + ez + 2V/(z + 9z + z#)(ab + be + ca) < a+b+c

Ukraina, 2001

Solution 1:

We will use Cauchy-Schwarz Inequality twice First, we can write ax + by +

cz < V2 + bỀ + c2 - +2 + y2 + z3 and then we apply again Cauchy-Schwarz In- equality to obtain

a + bụ +ez + 2V(zw + z + z#)(ab + be + ca) <

Vie Sia? + V2ab- V2” zụ <

The inequality being homogenuous in a,b,c we can assume that a+6+c= 1

We apply this time the AM-GM Inequality and we find that ax + by + cz + 2/ (zy + yz + zx) (ab + be + ca) < ax+byt+ez+ary+yz+z2+ab+be+ca But it follows

1—-z2—?®—z2 1-a2—b?- c?

the last one being clearly equivalent to ( — a)2 + ( — ð)? + (z — e)2 > 0

48 [| Calin Popa |] Let a,b,c be positive reals, with a,b,c € (0,1) such that ab+be+ ca =1 Prove that

1— tan? — an? 2 2 lan S =

& tan Atan Btan C(tan A+tan B+tan C) > 3(tan Atan B+tan BtanC+tanC tan A) S

clearly true because tan A,tan B,tanC > 0

49 | Titu Andreescu and Gabriel Dospinescu | Let #,,z < 1 and + + z = 1 Prove that

1+z2 1+2 14227 10

Solution:

Using the fact that (4 — 3t)(1 — 3#)? > 0 for any t < 1 we find that

127

1+2? — 50 Writing two other similar expressions for y and z and adding them up, we find the

In fact, this problem is equivalent to that difficult one Try to prove this!

50 [ Gabriel Dospinescu | Prove that if 2 + /y+ /z =1, then

(l—2)?(1—y)?(1 — 2)? > 2° ryz(x + y)(y + z)(ø + #)

holds

Solution:

We put a= /z, b= fy andc= Vz Then 1— #= 1— a2= (a+b+c)?— «2= (b+ c)(2a+6+c) Now we have to prove that ((a+ 6)(b+c)(e+a)(2a+b4+c)(a+ 26+ c)(a +b + 2c))? > 2'°a?b?c? (a? + b7)(b? + c?)(c? + a”) But this inequality is true as

b 4

it follows from the following true inequalities ab(a? + bŸ) < =3 (this equivalent

51 For any three positive real numbers a, b,c show that the following inequality holds

But this inequality is almost trivial Indeed, using the AM-GM Inequality we have

» z2z > 3 and so it remains to prove that » x? > » x, which follows from the inequalities

Further on

(cy +ez+yz—3)? = (cyt az+yz) —6(ay+az+yz)+9>

which is Chebyshev Inequality for the systems (21,%, ,%,) and

If one of x,y, z is negative, let us say x then

2+#z—#——z=(2—w—z)—z(1—z) >0 because y + z < 4⁄2(2 + z?) < 2 and zự < vty < 1 So we may assume 2, y, z are positive, let us consider 0< # < < z l z < 1 then

2+ #z— # —— z= (L— z)(1— z) + (L—z)(1— g) >0

Now if z > 1 we have

zt(@t+y) < V22 + (+ +)”) =2V1+ zy < 24+ ary < 2+ xyz

This ends the proof

Solution 2:

Using Cauchy-Schwarz Inequality, we find that

at+yt+z2—ayz=a(l—yz)t+ytz< W(œ2+(y+z)2)-(1+(1— g2))

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 > y? + z? > 2yz

55 Let x,y, z be positive real numbers such that xyz = x+y+2+2 Prove that (1) z +z + zz 3 2(z + +2);

with the analogous ones

56 [ Vasile Cârtoaje, Mircea Lascu | Let a, b,c be positive real numbers such that a2b2 + b2c2 + c2a2 = 3 Prove that

a+b+c>abc+2

Solution:

Assume, without loss of generality, that a > b > c, and rewrite the inequality in

a+b—2> (ab—1) (Sar) Let us denote z = Vab From 3a2b? > a2b? + b?c? + Ca? = 3, we have x > 1, and from a2b? < a2b2 + b2c2 + c2a2 = 3 we have z < \/3 Because a+b > 2Vab = 2x, and a* + b? > 2ab = 2x”, it suffices to prove that

— x4

2(z — 1) > (œ2 T— 1) (° =) & dx? > (2 +:1)(3— 2%)

The last inequality is true because 2x7 > 1 + z and 2> 3— zt

57 Find the maximum value of the expression

when a,b,c,d are reals whose sum of squares is 1

Solution:

The idea is to observe that a®(b+c+d)+b?(c+d+a)+c?(d+at+b)+d?(a+b+c)

is the same as » ab(a” + b*) Now, because the expression ab(a? + b2) appears when writing (a — b)*, let us see how the initial expression can be written

The maximum is attained fora =b=c=d= -—

58 Given are the positive numbers a,b,c and x,y,z, for which a+ a2 =b+y= c+2z=1 Prove the inequality

1 1 1 (abe + xyz)(— + ¬ >3

Proposed for the Balkan Mathematical Olympiad

it suffices to prove that

Sa (b? — be +c’) 224

This can be written as

ye IV À a(b + e)(b2 + c°) —3abe(at+bt+e) > at +abe(atb+e) >

Solution 2:

Rewrite our inequality as

` (b + e)aŸ ` 3(ab + be + ca)

But this follows from a more general result:

If a,b,c, x,y,z > 0 then

aly +2) ` a2.“ NẠP

_b+cS my: + But this inequality is an immediate (and weaker) consequence of the result from problem 109

So we get

A+B = (SN pa) =

from Cauchy-Schwarz Inequality So we get

A> 50), Vb +0? 20 a= > va+5- vB+ —Sea

We denote

Ag =Veta:-Vb+a-a= 22a

Svab+a?+a and the similar relations with A, and A, So we get A > A, + Ay + A- But because )›» a)? > 300 ab) we get

A¿> [Ee ai, a \V 3 dab - ee + va)

and also the similar inequalities are true So we only need to prove that

we get » \/ 3 + (— > 2 We have equality if and only ifa=b=c

60 [| Murray Klamkin ] Find the maximum value of the expression #7 + #23 +

co CC ay, +072, when 71, 2%2, ,%n-1,%n > 0 add up to 1 and n > 2

112 T 49a T''' Tđặ 1#n TTn41 Szc

27 for all #1,22, ,2Z„_-1,#„ > O which add 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 ro = min{2%1,%2, ,%n41} But this implies that

rit, +2503 + + ara, < (x, +22)?a3 + a3e4t - +07 18a + 2) (#1 + a3) 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+ 6?c+ 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 a?b + b2®e + e2a < (a + 3) - (0 + 5) follows immediately from abe > b’e, ae > ac Because

Let ———— = q; The problem reduces to proving that for any a; + ag +

1998 + z;

+ - + d„ = 1 we have the inequality

II(-)) >(m— 1)"

This inequality can be obtained by multiplying the obvious inequalities

62 Prove that for any a,b,c > 0 we have

(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 sides of a triangle ABC’ With the usual notations in a triangle, the inequality becomes

16K?

2 p21 2)

(a +6 +O) tae < abc(ab+be+ca) & (at+b+c)(ab+bc+ca)R? > abc(a* +b? +e’)

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

Inequality ¡t follows that ø!“° = (1+ ø— 1)? <1+(a—1)(1—-b)=a+b-—ab

and thus the conclusion Now, it x or y is at least 1, we are done Otherwise, let

0 < z,y < 1 In this case we apply the above observation and find that #¥ + y® >

x T—> x + =1

64 Prove that for any #1, ,#„,11,.-.,„ © R such that 27 + -+ 27 =

ys + + y2 = 1 we have the inequality (x, y2 — xy)" < 2 ( — Ys)

mens tw -nw? = (Set) (Su) - (Seam) =

Vi + fiz to + Vin <cVui Fant Fy

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 & Eor

example, we can take #„ = 2#, because in this case we have x1 +ao+: +a,% = ©p41—2 Thus, we find that

k=1

c>

for any n Taking the limit, we find that c > 1+ V2 Now, let us prove that 1 + V2

works We will prove the inequality

Jui + (ite + fin < (1+ V2I)Va Faye Ee

by induction For n = 1 or n = 2 it is trivial Suppose it is true for n and we will prove that v/#1 + 4⁄22 + - +4/#„ T /#a+i < (1 + X⁄2)⁄1 + đa 7+ -'': +#n T#n+1

Of course, it is enough to prove that

aaah te aah 2 Van tN

And so we are left with the inequality

3 3 +3Vabe —3

which is in fact an identity!

67 | Mircea Lascu and Vasile Cartoaje | For any a,b,c > 0 we have

(a? + ab + b)(b2 + be + e”)(c? + ca + a”) > (ab + be + ca)?

Solution 2:

Using the inequality

[[t+2 > ặ (2+) (Sa)

(+) > 3À `ab,

we have the following chain of inequalities

and the fact that

[](@ +0540?) > J] lúa + 0) = = [It + ĐỈ > (> ab) - (na) > ®S ab)

and the problem is solved