Show that for $x,y\in\mathbb{R}$ with $x,y\geq 0$, the arithmetic mean-quadratic mean inequality $$\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$$ holds.

After my calculations I'll get:

$$-x^2+2xy-y^2$$ which can't be $\leq 0$.


Completing the square gives:

$-x^2+2xy-y^2=-(x+y)^2\leq 0$


We can square both sides of your inequality, to get

$$\frac{x^2+2xy+y^2}{4} \leq \frac{x^2+y^2}{2}$$

$$\frac{x^2+2xy+y^2}{4} - \frac{x^2+y^2}{4} \leq \frac{x^2+y^2}{2}-\frac{x^2+y^2}{4} $$

$$\frac{xy}{2} \leq \frac{x^2+y^2}{4} $$

$$2xy \leq x^2+y^2 $$

$$0 \leq (x-y)^2$$

This inequality is true, since $x-y$ is a real number and all squares of real numbers are non-negative. We have shown that your inequality is equivalent to one that we know is always true. Therefore yours is always true.

  • 2
    $\begingroup$ It is not true that $a^2\geq a$ for all real $a$, but it doesn't matter. We can still justify squaring each side of the inequality. If the LHS is negative then the inequality obviously holds. Otherwise, each side is non-negative, and $f (x)=x^2$ is an increasing function for non-negative $x$. $\endgroup$ – Dylan Nov 2 '15 at 7:50
  • $\begingroup$ @Dylan Thanks for the correction - not sure why I wrote that, seems very silly in hindsight. $\endgroup$ – Zubin Mukerjee Nov 2 '15 at 9:41

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