<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] For $\sin{x}+1>0$ we obtain: $$2x^2\sin x+2x\cos x+2x^2+1=2(1+\sin{x})x^2+2x\cos{x}+1\geq0$$ because
$$\frac{\Delta}{4}=\cos^2x-2(1+\sin{x})=-(1+\sin{x})^2<0.$$
If $\sin{x}+1=0$ so $$2x^2\sin x+2x\cos x+2x^2+1=1>0.$$
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