prove that $x^4+y^4-4b^2xy \geq -2b^4$ for all $x,y \in \Bbb{R}$?

Question

How would you prove that $$x^4+y^4-4b^2xy \geq -2b^4\;\;\;\;\;\;\;\; \forall x,y \in \Bbb{R}?$$

My idea was to show that the function $H(x,y)=x^4+y^4-4b^2xy+2b^4 \geq 0$ for all $x,y \in \Bbb{R}$ and to show that I am considering finding the absolute minimum of the function and show that it is equal to or greater then $0$ but im strugging to show this

Answer 1

Since $x^4+y^4\geq 2x^2y^2$ we have:

$$x^4+y^4-4b^2xy +2b^4\geq 2x^2y^2 -4b^2xy +2b^4 =2 (xy-b^2)^2\geq 0$$