Proving inequality using Lagrange multipliers, somehow?

Question

While going over assignments preparing for an upcoming exam, I noticed the question

Prove that $x^{4} + y^{4} - 4b^{2}xy \geq -2b^{4} \text{ }\forall\text{ } x,y \in \mathbb{R}$

I had used the inequality $2|a||b| \leq a^{2} + b^{2}$ and a whole lot of manipulation to get the result, but it had been docked marks for not using methods taught in class. In class, this question was part of the chapter in which we covered the Lagrange algorithm for finding the max/min of a function given a certain constraint, as well as the algorithm to find the max/min of a closed and bounded function on $\mathbb{R}^{2}$.

I cannot figure out how to approach this problem using either one of those algorithms, any ideas?

Answer 1

Hints: let $f(x,y) = x^4+y^4-4b^2xy$. Show that the critical points (gradient equals zero) of $f$ are $(b,b)$ and $(-b,-b)$, show that they are local minima, and note that $f$ at these points takes the value $-2b^4$.

Answer 2

$ A = x^{4} + y^{4} - 4b^{2}xy + 2b^4$

$A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2x^{2}b^2 + 2y^2b^2 - 4b^{2}xy $

$A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x^2 + y^2 -2xy)$

$A = (x^{2} - b^{2})^2 + (y^{2} - b^{2})^2 + 2b^2(x - y)^2$

Which is positive as a sum of squares