Trigonometric inequality $\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}$ for all $x>1/2$

Question

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.

Answer 1

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 $x\geq\frac{1}{2}$, $$ \arctan\left(\frac{1}{1-x+x^2}\right) > \frac{1}{x^2+1}, \qquad \frac{1}{1-x+x^2}>\tan\left(\frac{1}{x^2+1}\right). $$

Answer 2

Taking arctangent of both sides (and using its monotonicity in a given range):

$$f_1(x)=\frac{1}{1+x^2} \\ f_2(x)=\arctan \frac{1}{1-x+x^2}$$

$$f_2(x)= \int_0^1 \frac{1-x+x^2}{(1-x+x^2)^2+t^2} dt$$

We need to prove that:

$$\int_0^1 \frac{(1-x+x^2)(1+x^2)}{(1-x+x^2)^2+t^2} dt >^? 1$$

$$\int_0^1 \frac{1-x+2x^2-x^3+x^4}{(1-2x+3x^2-2x^3+x^4)+t^2} dt >^? 1$$

$$1+\int_0^1 \frac{x(1-x+x^2)-t^2}{(1-x+x^2)^2+t^2} dt >^? 1$$

Now we need to prove that:

$$f_3(x)=\int_0^1 \frac{x(1-x+x^2)-t^2}{(1-x+x^2)^2+t^2} dt >^? 0, \quad x> \frac12$$

Since $1-x+x^2 \geq 1$ for $x \geq 1$, the inequality follows trivially in this case, so we only need to consider the range:

$$\frac12 <x<1$$

$$f_3'(x)=\int_0^1 \frac{(1-x^2)(1-x+x^2)^2-(1-4 x+3 x^2-4 x^3) t^2}{((1-x+x^2)^2+t^2)^2} dt$$

We have:

$$(1-x^2)(1-x+x^2)^2-(1-4 x+3 x^2-4 x^3)=x (2-x) (1+x^2)^2 >0, \quad 0<x<2$$

So $f_3(x)$ increases monotonely in the range of interest, which means it's enough to prove that:

$$f_3 \left( \frac{1}{2} \right) >^? 0$$

$$f_3 \left( \frac{1}{2} \right)=\int_0^1 \frac{\frac{3}{8}-t^2}{\frac{9}{16}+t^2} dt=\frac{3}{4} \int_0^{4/3} \frac{2/3-u^2}{1+u^2} du= \\ = \frac{5}{4} \arctan \frac{4}{3}-1 >0$$

This finishes the proof.

Note:

$$\arctan \frac{4}{3}= 2 \arctan \frac{3}{4} \left(\sqrt{1+\frac{16}{9}}-1 \right)=2 \arctan \frac{1}{2}$$

$$f_3 \left( \frac{1}{2} \right)=\frac{5}{2} \arctan \frac{1}{2}-1>\frac{5}{2} \left(\frac12-\frac{1}{24} \right)-1=\frac{7}{48}$$

Thus we don't need any numerical checks.

Answer 3

Let $f(x)=\arctan\frac{1}{1-x+x^2}-\frac{1}{1+x^2}.$

We need to prove that $f(x)>0$ for all $x>\frac{1}{2}.$

Indeed, $$f'(x)=\frac{(1-x)(3x+1)}{(1+x^2)^2(x^2-2x+2)},$$ which gives $f$ increases on $\left(\frac{1}{2},1\right]$ and decreases on $[1,+\infty)$ and since $\lim\limits_{x\rightarrow+\infty}f(x)=0,$ it's enough to prove that $$f\left(\frac{1}{2}\right)>0,$$ which is true.