<s>[INST] Prove the following trignometric inequality for all real $x$ Prove the following trignometric inequality for all $x \in \Bbb R$
$$x^2 \sin(x) + x \cos(x) + x^2 + {\frac 12} >0$$
take $x$  in the form of radians.
This particular question is the seventh question of the  1995 Indian RMO.
 [/INST] There may be more efficient ways but this is what I have:
If it is positive (i.e. nonnegative) then certainly when we multiply it by a positive function it will stay positive... to this end, multiply it by $\sin^2(x)\cos^2(x)$ this gives 
$$x^2\sin^3\cos^2(x)+x\sin^2(x)\cos^3(x)+x^2\sin^2(x)\cos^2(x)+\frac{1}{2}\sin^2(x)\cos^2(x)$$
Now this is positive provided that $x^2\cos^2(x)(\sin^2(x)-\sin^3(x))$ is positive and similarly if $\frac{1}{2}\cos^2(x)\sin^2(x) - x\sin^2(x)\cos^3(x)$ is positive.  I'll show that the first term is positive.  This first term is positive only if $g(x)=\sin^2(x)-\sin^3(x)$ is positive.
Upon taking the derivative we get that this function has roots precisely when $x=n\pi$ for $n\in \mathbb{Z}$. If we compute the second derivative we get
$$g''(x) = 3\sin^3(x)-2\sin^2(x)+2\cos^2(x)-6\sin(x)\cos^2(x)$$
and at $x=n\pi$ we have that $g''(x)=2>0$, therefore the function is always positive and likewise we can check the other term $\frac{1}{2}\cos^2(x)\sin^2(x) - x\sin^2(x)\cos^3(x)$ is positive.
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