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

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− | <math>(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2</math> with equality if and only if <math>\frac{xy^3}{z} = frac{yz^3}{x} = frac{zx^3}{y}</math>. 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. | + | <math>(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2</math> with equality if and only if <math>\frac{xy^3}{z} = \frac{yz^3}{x} =\frac{zx^3}{y}</math>. 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== | ==Solution 2== |

## Revision as of 19:37, 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 we can divide both sides of the inequality by , and .

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.