# work 026 – CRUX Mathematicorum

Problem

For $a,b$ positive reals prove that –

$\frac{a}{b} + \frac{b}{a} + \sqrt{1+\frac{2ab}{a^2+b^2}} \ge 2 + \sqrt{2}$

Proof

Let $a+b = 2p , a^2+b^2 = 2q^2 , ab = r^2$

then the inequality is equivalent to –

$\sqrt{2} \cdot q(q^2-r^2) \ge r^2 \cdot (q-p)$

Obviously we have that –

$\sqrt{2} \cdot q(q+r) \ge r^2$ (since $q \ge r$)

Also by AM-GM ,

$q-r \ge q-p$

multiplying these two we get the result with equality holding when

$q-r=q-p \implies a=b$