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.
 A: Put $t=\tan(\frac{x}{2})$. Now substitute it in the original expression. We have $$ x^{2}\sin x + x\cos x + x^{2} + \frac{1}{2}$$ $$= (1+\sin x)x^{2} + x\cos x+\frac{1}{2}$$ $$=(1 + \frac{2t}{1+t^{2}})x^{2} + x(\frac{1-t^{2}}{1+t^{2}}) + \frac{1}{2}$$ $$= \frac{(1+t)^{2}}{1+t^{2}}[x^{2} - x\frac{1-t^{2}}{(1+t)^{2}} + \frac{(1-t^{2})^{2}}{4(1+t)^{4}} + \frac{1+t^{2}}{4}]$$ $$= \frac{(1+t)^{2}}{1+t^{2}}[x-\frac{1-t^{2}}{2(1+t)^{2}}]^{2} + \frac{(1+t)^{2}}{4} > 0. $$ Hope it helps.
A: The discriminant of the "quadratic" equation is $$\cos^2x-2(\sin x+1)=-(\sin x+1)^2$$ and is non-positive.
When $\sin x=-1$, the inequation reduces to 
$$-x^2+x^2+1/2>0,$$ which is true.
When $\sin x\ne-1$, there cannot be real roots, and the function is positive everywhere.

Very interestingly, a plot of the function will lead you to believe in an infinity of zero values of the LHS, thus invalidating the claim. Anyway, all these are false roots.


A: Rewrite as $$f(x)=(1+\sin x)x^2+(\cos x) x+\frac 12$$
Multiply through by $4(1+\sin x)\ge 0$ to complete the square with $$4(1+\sin x)f(x)=$$
$$\left(2(1+\sin x)x+\cos x\right)^2-\cos^2x+2(1+\sin x)=(2(1+\sin x)x+\cos x)^2+(1+\sin x)^2\ge 0$$
As the sum of two squares (using $\cos^2 x=1-\sin^2 x$). Now any case of equality must have $1+\sin x=0$ and in this case $f(x)=\frac 12\gt 0$. So there is no case of equality for $f(x)$ and we have $f(x)\gt 0$.
A: 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.
