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

(→Solution) |
Anewpassword (talk | contribs) |
||

Line 62: | Line 62: | ||

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> | 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 6 === | ||

+ | We want to try and apply Cauchy (Titu), by transforming the numerator into a quadratic expression and the denominator into a linear expression. This is easily achieved by using the provided condition: \\ | ||

− | === Solution | + | Since <math>abc=1, \frac{1}{a^3(b+c)}=\frac{a^2b^2c^2}{a^3(b+c)}=\frac{b^2c^2}{a(b+c)}</math>. Likewise, <math>\frac{1}{b^3(c+a)}=\frac{a^2c^2}{b(a+c)},</math> and <math>\frac{1}{c^3(a+b)}=\frac{a^2b^2}{c(a+b)}</math>. Hence, by Cauchy (Titu): |

+ | |||

+ | \begin{align*} | ||

+ | \sum_{\text{cyc}}\frac{1}{a^3(b+c)} & \ge \frac{(bc+ac+ab)^2}{a(b+c)+b(a+c)+c(a+b)} \\& = \frac{(bc+ac+ab)^2}{2bc+2ac+2ab} \\& = \frac{bc+ac+ab}{2} \\& \ge \frac{3\sqrt[^3]{a^2b^2c^2}}{2}, \text{by AM-GM} \\& = \frac{3}{2}, \text{by the given condition <math>abc=1</math>}. | ||

+ | |||

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

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

## Revision as of 10:43, 20 June 2020

## 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

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 5

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 6

We want to try and apply Cauchy (Titu), by transforming the numerator into a quadratic expression and the denominator into a linear expression. This is easily achieved by using the provided condition: \\

Since . Likewise, and . Hence, by Cauchy (Titu):

\begin{align*}

\sum_{\text{cyc}}\frac{1}{a^3(b+c)} & \ge \frac{(bc+ac+ab)^2}{a(b+c)+b(a+c)+c(a+b)} \\& = \frac{(bc+ac+ab)^2}{2bc+2ac+2ab} \\& = \frac{bc+ac+ab}{2} \\& \ge \frac{3\sqrt[^3]{a^2b^2c^2}}{2}, \text{by AM-GM} \\& = \frac{3}{2}, \text{by the given condition }.

### 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.*