# 1983 IMO Problems/Problem 6

## Contents

## 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.

## Solution 3

Let be the semiperimeter, , of the triangle. Then, , , and . We let , and (Note that are all positive, since all sides must be shorter than the semiperimeter.) Then, we have , , and . Note that , so Plugging this into and doing some expanding and cancellation, we get The fact that each term on the left hand side has at least two variables multiplied motivates us to divide the inequality by , which we know is positive from earlier so we can maintain the sign of the inequality. This gives We move the negative terms to the right, giving We rewrite this as where is any real number. (This works because if we evaulate the cyclic sum, then as long as the coefficients of and on the right sum to 1 the right side will be .

## Solution4

"Solution from 111 problems in Algebra and Number Theory"