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

First,
$$
a^2+3b^2\geq 0
$$
Now,
$$
a^2+2ab+b^2+a^2-2ab+b^2\geq a^2-b^2
$$
Thus,
$$
(a+b)^2+(a-b)^2\geq (a+b)(a-b)
$$
Let $a+b=x, a-b=y$, i.e., $a=\frac{x+y}{2}, b=\frac{x-y}{2}$ which is one-to-one correspondence between $(x,y)$ and $(a,b)$, then,
$$
x^2+y^2\geq xy
$$
Q.E.D
