Show that $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0$ 
Show that for any real number $x$:
   $$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0.$$

$\bf{My\; Try::}$ Using $a\sin x+b\cos x\geq -\sqrt{a^2+b^2}$
So $$x^2\sin x+x\cos x\geq -\sqrt{x^4+x^2}=-x\sqrt{1+x^2}$$
and $$4x^4+4x^2+1>4x^4+4x^2\Rightarrow (2x^2+1)^2>4x^2(x^2+1)$$
So $$(2x^2+1)>2x\sqrt{x^2+1}\Rightarrow x^2+\frac{1}{2}>x\sqrt{x^2+1}$$
So $$x^2\sin x+x\cos x+x^2+\frac{1}{2}>-x\sqrt{x^2+1}+x\sqrt{x^2+1}=0$$
So $$x^2\sin x+x\cos x+x^2+\frac{1}{2}\gt 0\;\forall x\; \in \mathbb{R}$$
Is my solution is right, If not Then how can we solve it, Help required, Thanks in Advance

If we consider $a=(1+\sin x),\, b=\cos(x),\, c=\frac{1}{2}$ we have that
$$ b^2-4ac = \cos^2(x)-2-2\sin(x)=-4\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)^4 \leq 0$$
hence $ax^2+bx+c$ is never negative, since $a\geq 0$ and $\Delta=b^2-4ac\leq 0$.
