# 1964 IMO Problems/Problem 2

## Problem

Suppose $a, b, c$ are the sides of a triangle. Prove that

$$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}.$$

## Solution

Let $b+c-a = x$, $c+a-b = y$, and $a+b-c = z$. Then, $a = \frac{y+z}{2}$, $b = \frac{x+z}{2}$, and $c = \frac{x+y}{2}$. By AM-GM, $$\frac{x+y}{2} \geq \sqrt{xy},$$ $$\frac{y+z}{2} \geq \sqrt{yz},$$ $$\textrm{and }\frac{x+z}{2} \geq \sqrt{xz}.$$

Multiplying these equations, we have $$\frac{x+y}{2} \cdot \frac{y+z}{2} \cdot \frac{x+z}{2} \geq xyz$$ $$\therefore abc \geq (a+b-c)(b+c-a)(c+a-b).$$ We can now simplify: $$(a+b-c)(b+c-a)(c+a-b) \leq abc$$ $$(-a^2 + b^2 - c^2 + 2ac)(c+a-b) \leq abc$$ $$a(-a^2 + b^2 - c^2 + 2ac) + c(-a^2 + b^2 - c^2 + 2ac) - b(-a^2 + b^2 - c^2 + 2ac) \leq abc$$ $$-a^3 + ab^2 - ac^2 + 2a^2c - a^2c + b^2c - c^3 + 2ac^2 + a^2b - b^3 + bc^2 - 2abc \leq abc$$ $$a^2b + a^2c - a^3 + b^2c + ab^2 - b^3 + ac^2 + bc^2 - c^3 - 2abc \leq abc$$ $$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le{3abc}\textrm{. }\square$$ ~mathboy100

## Solution 2

We can use the substitution $a=x+y$, $b=x+z$, and $c=y+z$ to get

$$2z(x+y)^2+2y(x+z)^2+2x(y+z)^2\leq 3(x+y)(x+z)(y+z)$$

$2zx^2+2zy^2+2yx^2+2yz^2+2xy^2+2xz^2+12xyz\leq 3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz$

$$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\geq 6xyz$$

$$\frac{x^2y+x^2z+y^2x+y^2z+z^2x+z^2y}{6}\geq xyz=\sqrt[6]{x^2yx^2zy^2xy^2zz^2xz^2y}$$

This is true by AM-GM. We can work backwards to get that the original inequality is true.

## Solution 3

Rearrange to get $$a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b) \ge 0,$$ which is true by Schur's inequality.