# Difference between revisions of "1983 IMO Problems/Problem 6"

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Without loss of generality, let <math>a \geq b \geq c > 0</math>. By Muirhead or by AM-GM, we see that <math>a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>. | Without loss of generality, let <math>a \geq b \geq c > 0</math>. By Muirhead or by AM-GM, we see that <math>a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>. | ||

− | If we can show that <math>a^3 b + b^3 c+ c^3 a | + | If we can show that <math>a^3 b + b^3 c+ c^3 a \geq a^3 c + b^3 a + c^3 b</math>, we are done, since then <math>2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)</math>, and we can divide by <math>2</math>. |

We first see that, <math>(a^2 + ab + b^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>. | We first see that, <math>(a^2 + ab + b^2) \geq (b^2 + bc + c^2)</math>, so <math>(a-c)(b-c)(a^2 + ab + b^2) \geq (a-c)(b-c)(b^2 + bc + c^2)</math>. |

## Revision as of 20:00, 27 August 2017

## Problem 6

Let , and be the lengths of the sides of a triangle. Prove that

.

Determine when equality occurs.

## Solution 1

By Ravi substitution, let , , . Then, the triangle condition becomes . After some manipulation, the inequality becomes:

.

By Cauchy, we have:

with equality if and only if . So the inequality holds with equality if and only if x = y = z. Thus the original inequality has equality if and only if the triangle is equilateral.

## Solution 2

Without loss of generality, let . By Muirhead or by AM-GM, we see that .

If we can show that , we are done, since then , and we can divide by .

We first see that, , so .

Factoring, this becomes . This is the same as:

.

Expanding and refactoring, this is equal to . (This step makes more sense going backwards.)

Expanding this out, we have

,

which is the desired result.