$3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$

Question

Prove that $3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ for all $x,y\geq 0$.

Expanding, the inequality becomes $$3x^2+3xy+3y^2-6x\sqrt{xy}-6y\sqrt{xy}+6xy\geq x^2+xy+y^2$$ which is $$x^2+4xy+y^2\geq3\sqrt{xy}(x+y)$$

We can try using AM-GM: $$x^2+xy+xy\geq 3\sqrt[3]{x^4y^2}$$ This is close to the right-hand side but still different.

Answer 1

$x^2+xy+y^2 = (x+\sqrt{xy}+y)(x-\sqrt{xy}+y)$, so divide through and get

$3(x-\sqrt{xy}+y) \ge x+\sqrt{xy}+y$

$2(x+y) \ge 4\sqrt{xy}$

$\frac{x+y}{2} \ge \sqrt{xy}$

which is AM-GM.

Answer 2

$\Leftrightarrow (\sqrt{x}-\sqrt{y})^2(x-\sqrt{xy}+y)\geq 0$.