Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$ 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.
Thanks in advance!
 A: 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.
A: Just for the record, a "geometric" proof:

OBS: Which provides a nice idea that can be generalized to prove the Young's Inequality.
A: $0 \le \frac12 (|x|-|y|)^2 = \frac12 (|x|^2+|y|^2)-|x||y| = \frac12 (x^2+y^2)-|xy|$
A: 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 $d \ne 0$).
A: 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 $$x^2+y^2+2xy=(x+y)^2\ge 0\to -(x^2+y^2)\le 2xy$$
A: 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.
