<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] $$g(x)=\frac{(1+x)^2}{3+x^2}=1+2\frac{x-1}{3+x^2}$$
$$\text{$g$ is continuous for all real numbers and has a derivative} \\ \text{therefore we can find the max and min} \\ g'=2\frac{3-x^2+2x}{(3+x^2)^2}; \quad g'=0\Rightarrow (x-3)(x+1)=0 \\ \text{After that it's obvious that the range of $g$ is } [f(-1),f(3)]=[0,\frac{4}{3}] $$
$$\text{At $g(2)=1$ the $\tan$ function inside $f(g)$ for $\frac{\pi}{2}$ isn't defined}$$
$$f(g)=\tan\Big(g(x)\frac{\pi}{2}\Big)\cdot \tan\Big(g(-x)\frac{\pi}{2}\Big)$$
$$g\cdot\frac{\pi}{2} :\mathbb{R} \rightarrow[0,\frac{2}{3}\pi]$$ 
$$\text{And also, at }x\geq0,\;g(x)\cdot \frac{\pi}{2}\in[\frac{\pi}{6},\frac{2}{3}\pi] \\ \hspace{135px} g(-x)\cdot\frac{\pi}{2}\in[0,\frac{\pi}{2})$$
$$\text{And since when both $tan$ parts of the function when $g(x \geq 0)\cdot \frac{\pi}{2}\in (\frac{\pi}{2}, \frac{2}{3}\pi]$} \\ \text{are multiplied, we get a negative result,} \\ \text{Therefore we should check for $\forall x\;\;g(x\geq 0)\cdot \frac{\pi}{2}\in[\frac{\pi}{6}, \frac{\pi}{2})$}$$
$$\text{From here it should be simpler to continue, what you should end up} \\ \text{with is that at $\frac{\pi}{6}$ the $\tan$ functions take their maximal value when} \\ \text{$g(x)\frac{\pi}{2} = \frac{\pi}{6} \;$and multiplied together, the result will be $\frac{1}{3}$}$$
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