<s>[INST] Prove that $|a\sqrt{1-b^2}+b\sqrt{1-a^2}-\sqrt{3(1-a^2)(1-b^2)} +\sqrt{3}ab| \le2$ 
Prove for any $a, b \in [-1, 1]$ that $$|a\sqrt{1-b^2}+b\sqrt{1-a^2}-\sqrt{3(1-a^2)(1-b^2)} +\sqrt{3}ab| \le2$$

I'm sure there is a solution using the Cauchy-Swartz inequality. Thus i tried to prove $$\Big(a\sqrt{(1-b)(1+b)}+b\sqrt{(1-a)(1+a)}-\sqrt{3(1-a)(1+a)(1-b)(1+b)}+\sqrt{3}ab\Big)^2 \le 4$$
Can we use C-S to prove that?
 [/INST] Hint: if $a=\cos\alpha$, $\sqrt{1-a^2}=\sin\alpha$, $b=\cos\beta$ and $\sqrt{1-b^2}=\sin\beta$, then you need to show that
$$
\lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta|
=\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2,
$$
that is,
$$
\left|\frac12\sin(\alpha+\beta)-\frac{\sqrt3}2\cos(\alpha+\beta)\right|\le1.
$$
Of course, the last one is immediate.
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