<s>[INST] How to show that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is strictly increasing for $x \geq 1$? I am trying to prove that $f(x) = x \arctan \left(x \sin^{2} \left(\frac{1}{x}\right)\right)$ is a strictly increasing function for $x \geq 1$. 
I try to do this by showing that $f'(x)>0$ for all $x \geq 1$. 
We have $$f'(x)= \arctan(x \sin^{2} \frac{1}{x}) + \frac{x}{1+x^{2} \sin^{4} \frac{1}{x}} \left(\frac{-2\cos\frac{1}{x} \sin \frac {1}{x}}{x} + \sin^{2} \frac{1}{x}\right),$$
so it's not clear if this quantity is positive.
Now I can show that $x \arctan \left(\frac{1}{x}\right)$ is strictly increasing (by computing its derivative and using simple trig inequalities to show that it is positive), and that $$x \sin^{2} \left(\frac{1}{x}\right)- \frac{1}{x}>0$$ So my aim is somehow to compare $f(x)$ with $x\arctan \frac{1}{x}$ and use the above inequality to prove that $f'(x)>0$.
But now I am stuck. Any suggestions as to how to proceed? Or any other approach to the problem would be appreciated!
 [/INST] I have proved it as follows. Taking the expression for $f'(x)$, multiplying through by $(1+x^{2} \sin^{4} \frac{1}{x})$, we need to prove the following inequality, call it (a):
$$\left(1+x^{2} \sin^{4} \frac{1}{x}\right) \arctan(x \sin^{2} \frac{1}{x}) + x\sin^{2}\frac{1}{x} > 2\sin\frac{1}{x} \cos\frac{1}{x} \qquad \mbox{(a)}$$
Observe that since $2\sin\frac{1}{x} \cos \frac{1}{x} < 2x\sin \frac{1}{x}$ for $x>1$, if we prove that the L.H.S. side of (a) is greater than $2x \sin \frac{1}{x}$, we have automatically proved (a) itself.
Let $y = x \sin^{2}\frac{1}{x}$ and consider
$$g(y)=(1+y^{2}) \arctan y + y - 2y = (1+y^{2}) \arctan y - y$$
Now $g'(y) = 2y \arctan y>0$ if $y>0$ and thus $g(y)$ is strictly increasing; now $g(0)=0$ and hence $g(y)>0$ for $y>0$. 
Thus (a) is indeed true, and so $f'(x)>0$ for $x>1$. Proof is complete.
Any opinions on this?
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