How to prove $x^4+2x^2y^2+y^4\geq2xy^3$ Suppose that $x,y$ are real numbers. I want to prove $$x^4+2x^2y^2+y^4\geq2xy^3.$$ I noticed that this is the same as $$(x^2+y^2)^2\geq 2xy^3.$$ Can we proceed from here?

We can assume without loss of generality that $x,y\ge 0$ (why?)
So by AM-GM inequality:
$$x^2+y^2=x^2+\frac{y^2}3+ \frac{y^2}3+ \frac{y^2}3\ge4\sqrt[4]{\frac{1}{3^3}x^2y^6}=4\cdot3^{-\frac34}\sqrt{xy^3}.$$
By squaring:
$$(x^2+y^2)^2\geq16\cdot3^{-\frac32}xy^3\approx3.08 xy^3.$$
So we have the sharper bound $$\boxed{(x^2+y^2)^2\geq 3xy^3.}$$
