1983 IMO Problems/Problem 6

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Problem 6

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

$a^2 b(a-b) + b^2 c(b-c) + c^2 (c-a) \geq 0$.

Determine when equality occurs.

Solution 1

By Ravi substitution, let $a = y+z$, $b = z+x$, $c = x+y$. Then, the triangle condition becomes $x, y, z > 0$. After some manipulation, the inequality becomes:

$xy^3 + yz^3 + zx^3 \geq xyz(x+y+z)$.

By Cauchy, we have:

$(xy^3 + yz^3 + zx^3)(z+x+y) \geq xyz(y+z+x)^2$ with equality if and only if $\frac{xy^3}{z} = frac{yz^3}{x} = frac{zx^3}{y}$. 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.