# Simple Inequality Proof

Prove the following inequality

$\ xX + yY \leq \sqrt{ x^2 + y^2}\cdot\sqrt {X^2 + Y^2}$

where $x, y, X$ and $Y$ are real numbers.

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Hint: proof by contradiction. –  Newb Feb 3 '14 at 1:17

(2) Use the fact that $$2AB \leq A^2+ B^2$$
Use Cauchy-Schwarz Inequality for $\mathbb{R}^2$ with the "vectors" $(x,y)$ and $(X,Y)$.