<s>[INST] Trigonometric inequality $\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}$ for all $x>1/2$ In Hobson's book "A treatise on plane trigonometry", the following exercise is to be found

If $x>1/2$ then 
  $$\frac{1}{1+x+x^2}<\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}.$$

The first inequality is easy as it follows directly from $\alpha<\tan(\alpha)$ and the fact that $x$ is positive.
For the second one can produce the even tighter bound (on the given interval, checked only graphically)
$$\tan(\alpha)<\frac{\alpha}{1-\frac{\alpha^2}{2}}$$
but from here I cannot get to the exact given inequality.
I am aware of the fact that $\tan(\alpha)\sim \alpha$ for small $\alpha$, but I cannot manage to find the exact bound given in the exercise.
Any hints are welcome.
 [/INST] The given inequality is equivalent to 
$$ \frac{1}{1-\sqrt{z-z^2}}-1\geq\frac{\tan(z)}{z}-1\geq\frac{1}{1+\sqrt{z-z^2}}-1 $$
for $z\in\left(0,\frac{4}{5}\right]$. This is fairly simple to prove by concavity/convexity: the LHS lies above the line $y=\frac{z}{2}$, the central term lies below it. The central term lies above the line $y=0$, the RHS lies below it.
Here it is a simpler derivation: by Lagrange's theorem
$$ \arctan\left(\frac{1}{1+x+x^2}\right)=\arctan(x+1)-\arctan(x)=\frac{1}{\xi^2+1},\quad \xi\in(x,x+1) $$
hence
$$ \arctan\left(\frac{1}{1+x+x^2}\right) < \frac{1}{x^2+1}, \qquad \frac{1}{1+x+x^2}<\tan\left(\frac{1}{x^2+1}\right). $$
Similarly
$$ \arctan\left(\frac{1}{1-x+x^2}\right)=\arctan(x)-\arctan(x-1)=\frac{1}{\eta^2+1},\quad \eta\in(x-1,x) $$
hence, assuming $x\geq\frac{1}{2}$,
$$ \arctan\left(\frac{1}{1-x+x^2}\right) > \frac{1}{x^2+1}, \qquad \frac{1}{1-x+x^2}>\tan\left(\frac{1}{x^2+1}\right). $$
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