# 2004 JBMO Problems/Problem 1

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Prove that the inequality $$\frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } }$$ holds for all real numbers $x$ and $y$, not both equal to 0.

## Solution

Since the inequality is homogeneous, we can assume WLOG that xy = 1.

Now, substituting $m = (x+y)^2$, we have:

$m = x^2 + y^2 + 2xy = x^2 + y^2 + 2$ $\geq 2\sqrt {xy} + 2 = 4$, thus we have $m \geq 4$

Now squaring both sides of the inequality, we get: $$\frac{m}{(m-3)^2 } \leq \frac{8}{m-2}$$ after cross multiplication and simplification we get: $7m^2 -46m + 72 \geq 0$ or, $7(m-4)^2 +10(m-4) \geq 0$ which is always true since $m \geq 4$.

$Kris17$

## Solution 2

Again, since the inequality is homogenous, we can assume WLOG that $x^2+y^2=8$.

By AM-GM we gave $xy\leq4=\frac{x^2+y^2}{2}$ and by QM-AM we have that $x+y\leq4=2\sqrt{\frac{x^2+y^2}{2}}$.

Substituting we have $$\frac{x+y}{x^2-xy+y^2}\leq\frac{4}{4}=\frac{2\sqrt{2}}{\sqrt{8}}=\frac{2\sqrt{2}}{\sqrt{x^2+y^2}}$$

$DVDTSB$

## Solution 3

By Trivial Inequality, $$(x - y)^2 \geq 0 \iff 2(x^2 - xy + y^2) \geq x^2 + y^2 \iff \frac{2}{x^2 + y^2} \geq \frac{1}{x^2 - xy + y^2}.$$

Then by multiplying by $x + y$ on both sides, we use the Trivial Inequality again to obtain $2(x^2 + y^2) \geq (x + y)^2$ which means $$\frac{x+y}{x^2 - xy + y^2} \leq \frac{2(x + y)}{x^2 + y^2} \leq \frac{2\sqrt{2(x^2+y^2)}}{x^2 + y^2}$$ which after simplifying, proves the problem.