# Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$ [duplicate]

How can I prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$

I have tried substitute $x,y$ for numbers, which turns out right, but I don't know how to reason here.

Note that we have the following truth:

$$x^2+y^2-2xy=(x-y)^2\ge 0\to 2xy\le x^2+y^2$$ and $$x^2+y^2+2xy=(x+y)^2\ge 0\to -(x^2+y^2)\le 2xy$$

Nothing new, just with a slightly different approach: for $\,x,y\in\Bbb R$ :

$$\pm xy\le\frac{x^2+y^2}2\iff x^2\pm 2xy+y^2\ge 0\iff (x\pm y)^2\ge0$$

and since the rightmost inequality is trivial we're done.

• Good way of handling $x$ or $y$ being negative. – marty cohen Aug 18 '13 at 17:12

Just for the record, a "geometric" proof: OBS: Which provides a nice idea that can be generalized to prove the Young's Inequality.

$0 \le \frac12 (|x|-|y|)^2 = \frac12 (|x|^2+|y|^2)-|x||y| = \frac12 (x^2+y^2)-|xy|$

Trying a different way to prove $|xy|\leq\frac{x^2+y^2}{2}$.

If $x=y$, this becomes $x^2 \le x^2$ which is true.

If any of $x$ or $y$ are negative, let $v = |x|$ and $w = |y|$. The left side is $vw$ and the right side is $(x^2+y^2)/2 =(v^2+w^2)/2$ which is the same inequality with all variables non-negative. We can therefore assume that all variables are non-negative. $|xy|$ can then be replaced by $xy$.

If $x \ne y$, let $y = x+d$ where $d \ne 0$. This becomes $x(x+d) \le (x^2+(x+d)^2)/2 =x^2+xd+d^2/2$ or $x^2+xd \le x^2+xd+d^2/2$, which is obviously true (actually true with $<$ instead of $\le$ since $d \ne 0$).

I like to treat absolute values squaring.

I mean, like both members are positive, we have these equivalencies:

$$|xy| \le \frac12(x^2+y^2)$$ $$|xy|^2 \le \frac14 (x^2+y^2)^2$$ $$4x^2y^2 \le x^4+y^4+2x^2y^2$$ $$0 \le x^4+y^4-2x^2y^2$$ $$0 \le (x^2-y^2)^2$$

And the last one is clearly true, so it implies the first one.