# Simple inequality proof? [closed]

How do I show this inequality:

For positive real numbers $a,b$ such that $a+b=1$ then $\frac{2}{\frac{a}{x}+ \frac{b}{y}} \leq ax + by$ for $x, y > 0$

## closed as off-topic by Macavity, graydad, user147263, Najib Idrissi, apnortonFeb 28 '15 at 19:40

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• Are you sure the LHS has numerator $2$ and not $1$? Your inequality does not hold for say $a=b=x=y=\frac12$. – Macavity Feb 28 '15 at 16:17

The $2$ in your inequality should be $1$. Indeed,
$$(ax + by)\left(\frac{a}{x} + \frac{b}{y}\right) = a^2 + b^2 + ab\left(\frac{x}{y} + \frac{y}{x}\right) \ge a^2 + b^2 + 2ab\sqrt{\frac{x}{y}\frac{y}{x}} = a^2 + b^2 + 2ab = (a + b)^2 = 1,$$
with equality if and only if $x = y$.