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

$x^2+xy+y^2 = (x+\sqrt{xy}+y)(x-\sqrt{xy}+y)$, 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.
