<s>[INST] How can I show this inequality: $-2 \le \cos \theta (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$ 
Show that $$-2 \le \cos \theta ~ (\sin \theta +\sqrt{\sin ^2 \theta +3})\le 2$$ for all value of $\theta$.

Trial: I know that $0\le \sin^2 \theta \le1 $. So, I have $\sqrt3 \le \sqrt{\sin ^2 \theta +3} \le 2 $. After that I am unable to solve the problem. 
 [/INST] My Solution::  Given $$f(\theta) = \cos \theta \left(\sin \theta + \sqrt{\sin^2 \theta + 3}\right)$$
Now let $$y=\sin \theta \cdot \cos \theta  +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}$$
Now using the Cauchy-Schwarz inequality, we get
$$\left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \left(\sqrt{\sin^2 \theta + 3}\right)^2\right\}\geq \left\{\sin \theta \cdot \cos \theta +\cos \theta \cdot \sqrt{\sin^2 \theta + 3}\right\}^2$$
So we get $$y^2 \leq \left(\sin^2 \theta +\cos^2 \theta \right)\cdot \left\{\cos^2 \theta + \sin^2 \theta + 3\right\}=2^2$$
So we get $$-2 \leq y\leq 2\Rightarrow y\in \left[-2,2\right]$$
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