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 ?
 A: 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.
A: $$a^4+b^4+a^2b^2=(a^2+b^2)^2-a^2b^2=60$$
$$(a^2+b^2-ab)(a^2+b^2+ab)=60$$
Solving equations $$a^2+b^2-ab=6$$$$a^2+b^2+ab=10$$ will give you the result.
A: We have $$[4(a^2+b^2)]^2=16(a^4+b^4+2a^2b^2) =16(60+a^2b^2)$$ 


Then,  $$  [4(a^2+b^2)]^2 - (30 +ab)^2 =16(60+a^2b^2) - (30 +ab)^2 \\=15a^2b^2 -60 ab - 60 =\color{red}{15(ab-2)^2\ge0} $$ 
A: Let $a^2+b^2=2kab$.
Hence, we need to prove that
$$(4a^2+4b^2-ab)^2\geq900\cdot\frac{a^4+b^4+a^2b^2}{60}$$ or
$$(8k-1)^2\geq15(4k^2-1)$$ or
$$(k-2)^2\geq0.$$
Done!
A: $$4\sqrt{60+a^2b^2}-ab \geq 30$$
Let $t=ab$, so $$ 16(60+t^2)\geq (30+t)^2$$
$$ t^2-4t+4\geq 0$$ 
