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

Let us check for  $$x^2(1+\sin x)+x\cos x-y=0$$
As $x$ is real, the discriminant  $$\cos^2x+4(1+\sin x)y\ge0$$
Now $1+\sin x\ge0$
Check what if $1+\sin x=0?$
Else $4y\ge\sin x-1\ge-1-(-1)$
Observe that the equality cannot occur as $1+\sin x\ne0$
