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

Technique 1:$$\sqrt{x^2y^2} = xy \le \dfrac{x^2 + y^2}{2}\le x^2 + y^2$$ Technique 2:$$(x - y)^2 \ge 0 \implies x^2 +y^2 - 2xy \ge 0 \implies x^2 + y^2 \ge 2xy\ge xy$$ Technique 3 (my favorite): There's a statement $\dfrac{y}{x} + \dfrac{x}{y} \ge 2$ with many classical proofs (which I'd not state here). We can write the inequality as follows:$$\dfrac{y}{x}+\dfrac{x}{y} \ge 1$$Divide both sides by $xy$.$$\dfrac{1}{x^2} + \dfrac{1}{y^2} \ge \dfrac{1}{xy}$$Rewrite.$$\dfrac{x^2 + y^2}{x^2y^2} \ge \dfrac{xy}{x^2y^2}$$And finally...$$x^2 + y^2 \ge xy$$
