<s>[INST] Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verify that $f(x)$ is even and
$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$
which implies $f(x) \in (-\infty,\frac13]$, i.e.
$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$
and is visually confirmed below

However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.
So, the question is how to prove the inequality  $f(x) \le \frac13$ with rigor.
Note that it is equivalent to proving
$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$
 [/INST] Partial answer :
Well we can use Jensen's inequality remarking that :
$$f(x)=\ln\Big(\tan\Big(\frac{\pi}{2}\frac{(1+x)^2}{3+x^2}\Big)\Big)$$
Is concave on $[\frac{-17}{100},\frac{17}{100}]$
So we have :
$$f(x)+f(-x)\leq 2f(0)$$
A bit of algebra and we get the result !
We can improve the reasoning for that we use a substitution:
$$x=\frac{a}{a+1}$$
And follow the same reasoning !
The concavity (an attempt)for $-\frac{17}{100}<x<0$:
We start by introduce a function called $j$ :
$$j(x)=\ln\left(\tan\left(\frac{\left(\frac{x}{x+1}-1\right)^2}{3+\left(\frac{x}{x+1}\right)^2}\right)\right)$$
The derivative is equal to :
$$j'(x)=-\frac{\pi(4x+3)\csc\left(\frac{\pi}{2(4x^2+6x+3)}\right)\sec\left(\frac{\pi}{2(4x^2+6x+3)}\right)}{(4x^2+6x+3)^2}$$
Using the substitution $y=4x^2+6x+3$
The function :
$$h(y)=-\frac{\csc\left(\frac{\pi}{2(y)}\right)\sec\left(\frac{\pi}{2(y)}\right)}{(y)}$$
Is increasing and negative .On the other hand the function :
$$g(x)=\frac{\pi(4x+3)}{(4x^2+6x+3)}$$ is decreasing positive .
So as a multiplication we deduce that the function is negative increasing so the second derivative is positive . $j$ is convex .
We have $j\left(\frac{x}{1-x}\right)=f(-x)$
We Differentiate  and using the fact that :
$k(x)=\frac{-1}{(1-x)^2}$ is negative decreasing and $j$ is negative decreasing  we deduce by multiplication that $-f'(-x)$ is decreasing so $f(-x)$ is concave on the interval .
Hope it helps you !
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