1983 IMO Problems/Problem 6

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.

Solution 2

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

If we can show that $a^3 b + b^3 c+ c^3 a \geq a^3 c + b^3 a + c^3 b$ , we are done, since then $2(a^3 b + b^3 c+ c^3 a ) \geq a^3 b + a^3 c + b^3 c + b^3 a + c^3 a + c^3 b \geq 2(a^2 b^2 + a^2 c^2 + b^2 c^2)$ , and we can divide by $2$ .