<s>[INST] Algebraic and Trigonometric expression is $>0$ for all real $x$ 
Prove that $$2x^2\sin x+2x\cos x+2x^2+1$$ is always positive for all real $x$.

From completing the square method
Write $1$ as $\sin^2 x+\cos^2 x$
$$x^4+2x^2\sin x+\sin^2 x+x^2+2x\cos x+\cos^2 x+x^2-x^4$$
$$(x^2+\sin x)^2+(x+\cos x)^2+x^2(1-x^2)$$
$\bullet\; $ In $|x|\leq 1$ our expression is $>0$.
Is there is any way to prove expression $>0$ for $|x|>1$?
Thanks in advance.
 [/INST] By completing the square,
$$2(\sin x+1)x^2+2\cos x\,x+1\\
=2(\sin x+1)\left(x+\frac{\cos x}{2(\sin x+1)}\right)^2+1-\frac{\cos^2x}{2(\sin x+1)}\\
=2(\sin x+1)\left(\left(x+\frac{\cos x}{2(\sin x+1)}\right)^2+\frac{\sin x+1}2\right).$$
Obviously, $\sin x+1\ge0$. In case of equality, $\sin x=-1\implies \cos x=0$ and the expression reduces to $1$.
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