# Difference between revisions of "1995 IMO Problems/Problem 2"

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The desired conclusion follows. <math>\blacksquare</math> | The desired conclusion follows. <math>\blacksquare</math> | ||

+ | === Solution 3 === | ||

+ | Without clever substitutions: | ||

+ | By Cauchy-Schwarz, <cmath>\left(\sum_{cyc}\dfrac{1}{a^3 (b+c)}\right)\left(\sum_{cyc}a(b+c)\right)\geq \left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right) ^2=(ab+ac+bc)^2</cmath> Dividing by <math>2(ab+bc+ac)</math> gives <cmath>\dfrac{1}{a^3 (b+c)}+\dfrac{1}{b^3 (a+c)}+\dfrac{1}{c^3 (a+b)}\geq \dfrac{1}{2}(ab+bc+ac)\geq \dfrac{3}{2}</cmath> by AM-GM. | ||

+ | === Solution 3b === | ||

+ | Without clever notation: | ||

+ | By Cauchy-Schwarz, <cmath>\left(a(b+c) + b(c+a) + c(a+b)\right) \cdot \left(\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)}\right)</cmath> | ||

+ | <cmath>\ge \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^2</cmath> | ||

+ | <cmath>= (ab + bc + ac)^2</cmath> | ||

+ | |||

+ | Dividing by <math>2(ab + bc + ac)</math> and noting that <math>ab + bc + ac \ge 3(a^2b^2c^2)^{\frac{1}{3}} = 3</math> by AM-GM gives | ||

+ | <cmath>\frac{1}{a^3 (b+c)} + \frac{1}{b^3 (c+a)} + \frac{1}{c^3 (a+b)} \ge \frac{ab + bc + ac}{2} \ge \frac{3}{2},</cmath> | ||

+ | as desired. | ||

+ | |||

+ | === Solution 4 === | ||

+ | After the setting <math>a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z},</math> and as <math>abc=1</math> so <math>\left(\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1} {c}=1\right)</math> concluding <math>x y z=1 .</math> | ||

+ | |||

+ | <math>\textsf{Claim}:</math> | ||

+ | <cmath> | ||

+ | \frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | ||

+ | </cmath> | ||

+ | By Titu Lemma, | ||

+ | <cmath> | ||

+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)^{2}}{2(x+y+z)} | ||

+ | </cmath> | ||

+ | <cmath> | ||

+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)}{2} | ||

+ | </cmath> | ||

+ | Now by AM-GM we know that <cmath> (x+y+z)\geq3\sqrt[3]{xyz} | ||

+ | </cmath>and <math>xyz=1</math> which concludes to <math>\implies (x+y+z)\geq3\sqrt[3]{1}</math> | ||

+ | |||

+ | Therefore we get | ||

+ | |||

+ | <cmath> | ||

+ | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | ||

+ | </cmath>Hence our claim is proved ~~ Aritra12 | ||

+ | |||

+ | === Solution 5 === | ||

+ | Proceed as in Solution 1, to arrive at the equivalent inequality | ||

+ | <cmath> \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{3}{2} . </cmath> | ||

+ | But we know that <cmath>x + y + z \ge 3xyz \ge 3</cmath> by AM-GM. Furthermore, | ||

+ | <cmath> (x + y + y + z + x + z) (\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y}) \ge (x + y + z)^2 </cmath> | ||

+ | by Cauchy-Schwarz, and so dividing by <math>2(x + y + z)</math> gives | ||

+ | <cmath> \begin{align*}\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} &\ge \frac{(x + y + z)}{2} \\ &\ge \frac{3}{2} \end{align*},</cmath> | ||

+ | as desired. | ||

+ | |||

+ | === Solution 6 === | ||

+ | Without clever substitutions, and only AM-GM! | ||

+ | |||

+ | Note that <math>abc = 1 \implies a = \frac{1}{bc}</math>. The cyclic sum becomes <math>\sum_{cyc}\frac{(bc)^3}{b + c}</math>. Note that by AM-GM, the cyclic sum is greater than or equal to <math>3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math>. We now see that we have the three so we must be on the right path. We now only need to show that <math>\frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^\frac13</math>. Notice that by AM-GM, <math>a + b \geq 2\sqrt{ab}</math>, <math>b + c \geq 2\sqrt{bc}</math>, and <math>a + c \geq 2\sqrt{ac}</math>. Thus, we see that <math>(a+b)(b+c)(a+c) \geq 8</math>, concluding that <math>\sum_{cyc} \frac{(bc)^3}{b+c} \geq \frac32 \geq 3\left(\frac{1}{(a+b)(b+c)(a+c)}\right)^{\frac13}</math> | ||

+ | |||

+ | |||

+ | === Solution 7 from Brilliant Wiki (Muirheads) ==== | ||

+ | https://brilliant.org/wiki/muirhead-inequality/ | ||

+ | |||

+ | Scroll all the way down | ||

{{alternate solutions}} | {{alternate solutions}} | ||

## Latest revision as of 16:30, 19 June 2021

## Contents

## Problem

(*Nazar Agakhanov, Russia*)
Let be positive real numbers such that . Prove that

## Solution

### Solution 1

We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.

### Solution 2

We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.

### Solution 3

Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.

### Solution 3b

Without clever notation: By Cauchy-Schwarz,

Dividing by and noting that by AM-GM gives as desired.

### Solution 4

After the setting and as so concluding

By Titu Lemma, Now by AM-GM we know that and which concludes to

Therefore we get

Hence our claim is proved ~~ Aritra12

### Solution 5

Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.

### Solution 6

Without clever substitutions, and only AM-GM!

Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that

### Solution 7 from Brilliant Wiki (Muirheads) =

https://brilliant.org/wiki/muirhead-inequality/

Scroll all the way down
*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*