How can I prove that $xy\leq x^2+y^2$? [closed]

Solution 1:

$$x^2+y^2-xy=\frac{x^2}{2}+\frac{y^2}{2}+\frac{(x-y)^2}{2}$$

Solution 2:

Use polar coordinates: $$x=r \cos \theta,y=r \sin \theta $$

Your inequality becomes $$r^2 \cos \theta \sin \theta \leq r^2 $$

which is pretty much trivial.

Solution 3:

Though there are quite a few proofs already given, I'd like to add a visual one.

Solution 4:

\begin{align} 0\leq (x-y)^2 \implies & 0\leq x^2-2xy +y^2 \\ \implies & 2xy\leq x^2+y^2 \\ \implies & xy\leq \dfrac{x^2+y^2}{2} \\ \implies & xy\leq {x^2+y^2} \end{align} since the clearly nonnegative real $x^2+y^2$ clearly satisfies $\dfrac{x^2+y^2}{2} \leq x^2+y^2$.