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

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