Hello, it is $$\frac{(a-b)^2(a+b)^2(a^2+b^2-c^2)}{2(a^2+b^2)abc^2}\geq 0$$Sonnhard.

Pinko wrote: Prove that for each $\Delta ABC$ with an acute $\angle C$ the following inequality is true: $(a^2+b^2) cos(\alpha -\beta )\leq 2ab$. Let $A$ and $B$ be two distinct fixed points from plane $\alpha.$ Determine the locus of the points $C\in\alpha,$ $C\notin AB,$ for which $\angle{C}\leq\frac{\pi}2$ and $(a^2+b^2)\cdot\cos{(A-B)}=2ab.$

Pinko wrote: Prove that for each $\Delta ABC$ with an acute $\angle C$ the following inequality is true: $(a^2+b^2) cos(\alpha -\beta )\leq 2ab$. See also here: https://artofproblemsolving.com/community/c6h293296