Algebraic and Trigonometric expression is $>0$ for all real $x$

Question

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.

Answer 1

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 $$2x^2\sin x+2x\cos x+2x^2+1=1>0.$$

Answer 2

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 $1$.