How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?

$$x^2+y^2-xy=\frac{(2x-y)^2+3y^2}4=\frac{(2x-y)^2+(\sqrt3y)^2}4$$
Now, the square of any real numbers is $\ge0$
So, $(2x-y)^2+(\sqrt3y)^2\ge0,$ the equality occurs if each  $=0$
