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!
 A: The function is
$$
f(x)=x\tan^{-1}\left(x \sin^2\left(\tfrac{1}{x}\right)\right)
$$
For $x>1$, we have $x$ is stricly increasing and $\tan^{-1}\left(g(x)\right)$ preserves the behaviour of $g(x)$ because the arctangent function is monotonic.
So we can study the behaviour of $f(x)$ after sudying $g(x)$ for $x\in [1,\infty)$. After the change $u=\frac{1}{x}$, we have can study $g(1/u)=\tilde g(u)=\frac{\sin^2 u}{u}$ for $u\in (0,1]$. 
The first derivative of $\tilde g(u)$ is
$$
\tilde g'(u)=\frac{(2 u \cos u-\sin u) \sin u}{u^2}=\frac{\sin^2 u}{u^2}(2u \cot u-1)
$$
For $u\in(0,1]$ we have $\tilde g'(u)> 0$ because $2u>\tan u$ and then $2u \cot u-1>0$.
Thus $\tilde g'(u)$ is stricty increasing for $u\in(0,1]$ and then $g(x)$ is stricly decreasing for $x\in [1,\infty)$ and also $\tan^{-1}(g(x))$.
Now $f(x)$ is strictly increasing for $x\ge 1$, iff for $x_1,\,x_2\in [1,\infty)$
$$
x_2>x_1\quad \Longrightarrow \quad f(x_2)>f(x_1)
$$
that is $$x_2\tan^{-1}(g(x_2))>x_2\tan^{-1}(g(x_2)) \Leftrightarrow \frac{\tan^{-1}(g(x_1))}{\tan^{-1}(g(x_2))}>\frac{x_2}{x_1}>1 $$ that is $f(x)$ is strictly increasing for $x\ge 1$, iff $\tan^{-1}(g(x_2))<\tan^{-1}(g(x_1))$ and as we proved $\tan^{-1}(g(x))$ is strictly decreasing.
So we've proved that $f(x)$ is strictly increasing for $x\ge 1$. We have also that 
$$
\lim_{x\to\infty}f(x)=\lim_{u\to 0} \frac{\tan^{-1}(\tilde g(u))}{u}=1
$$
because for $u\to 0$, $\tan^{-1}(\tilde g(u))\sim \tilde g(u)=\frac{\sin^2 u}{u}$ and then $\frac{\tan^{-1}(\tilde g(u))}{u}\sim \frac{\sin^2 u}{u^2}\to 1$.
A: Simple, use derivative, not approximations. You get: 
$(x\arctan(x\sin^2(\frac{1}{x})))'=\arctan(x\sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(sin^2(\frac{1}{x}) + \frac{x}{1+x^2\sin^4(\frac{1}{x})}(-2x\sin\frac{1}{x}\cos\frac{1}{x}(-\frac{1}{x^2}))$ 
so all are positive for $x \geq 1$.
A: 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?
