Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$

My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+ab \leq210$$
Now what ?

we have to prove that $$a^2+b^2\geq \frac{15}{2}+\frac{ab}{4}$$
if $$\frac{15}{2}+\frac{ab}{4}<0$$ then is nothing to prove, in the other case we have
$$a^4+b^4+2a^2b^2\geq \frac{225}{4}+\frac{a^2b^2}{16}+\frac{15}{4}ab$$ for $$a^4+b^4$$ we Substitute $60-a^2b^2$ and we have to prove
$$60-a^2b^2+2a^2b^2\geq \frac{225}{4}+\frac{a^2b^2}{16}+\frac{15}{4}ab$$
rearranging and simplifying we get
$$a^2b^2-4ab+4\geq 0$$ which is $$(ab-2)^2\geq 0$$ which is true.
