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

$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right) \cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13 $$

Solution 1:

Note: $f(1) \triangleq \lim_{x\to 1} \tan ( \frac\pi2 \frac{(1+x)^2}{3+x^2}) \tan ( \frac\pi2 \frac{(1-x)^2}{3+x^2}) = 0$.

Since $f(x)$ is even, we only need to prove the case $x \in [0, \infty)$. Clearly $f(x) < 0$ for all $x > 1$. Also, $f(1) = 0$. Thus, we only need to prove the case $0\le x < 1$.

(Inspired by @pisoir's 1st equation) Let $A = \frac\pi2 \frac{(1+x)^2}{3+x^2}, B = \frac\pi2 \frac{(1-x)^2}{3+x^2}$. Clearly $A, B \in (0, \frac{\pi}{2})$. It suffices to prove that $\sin A \sin B \le \frac{1}{3}\cos A \cos B$ or $$\frac{1}{2}[\cos (A - B) - \cos ( A + B)] \le \frac{1}{3}\cdot \frac{1}{2}[\cos (A - B) + \cos (A + B)]$$ or $$\cos \frac{2\pi x}{x^2 + 3} \le 2 \cos \frac{\pi (x^2 + 1)}{x^2 + 3}. $$

Let $C = \frac{2\pi x}{x^2 + 3}$ and $D = \frac{\pi (x^2 + 1)}{x^2 + 3}$.

We split into two cases:

1. $\frac{1}{3} \le x < 1$:

Clearly, $\frac{\pi}{3} \le D \le \frac{\pi}{2}$ and $0 \le \frac{\pi}{2} - C \le 2(\frac{\pi}{2} - D) \le \frac{\pi}{2}$. We have \begin{align} \cos C &= \sin (\tfrac{\pi}{2} - C) \\ &\le \sin [2(\tfrac{\pi}{2} - D)]\\ &= 2\sin (\tfrac{\pi}{2} - D) \cos (\tfrac{\pi}{2} - D) \\ &\le 2\sin (\tfrac{\pi}{2} - D) \\ &= 2\cos D. \end{align}

2. $0 \le x < \frac{1}{3}$:

We give the following auxiliary results (Facts 1-2). The proofs are easy and thus omitted.

Fact 1: $\cos u \le 1 - \frac{1}{3}u^2$ for all $u$ in $[0, \pi/4]$.

Fact 2: $\cos v \ge \frac{1}{2} - \frac{5}{9}\sqrt{3}\, (v - \tfrac{\pi}{3})$ for all $v \in [\pi/3, \pi/2]$.

Let us proceed. Clearly $C \in [0, \pi/4]$ and $D \in [\pi/3, \pi/2]$. By Facts 1-2, it suffices to prove that $$1 - \frac{1}{3}C^2 \le 2\left[\frac{1}{2} - \frac{5}{9}\sqrt{3}\, \left(D - \frac{\pi}{3}\right) \right]$$ that is $$\frac{4\pi x^2 [9\pi - 5\sqrt{3}(x^2 + 3)]}{27(x^2 + 3)^2}\ge 0$$ which is true.

We are done.

Solution 2:

The problem is not too hard if you consider $$f(x)=\tan (A(x))\,\tan (B(x))$$ Using logarithmic differentiation $$\frac{f'(x)}{f(x)}=A'(x) \csc (A(x)) \sec (A(x))+B'(x) \csc (B(x)) \sec (B(x))$$

This gives $f'(0)=0$.

Repeating the process knowing that $f(0)=\frac 13$ and $f'(0)=0$ (this simplifies a lot the calculations), we have $$f''(0)=\frac{16}{81} \left(\sqrt{3}-\pi \right) \pi <0$$ So, by the second derivative test $x=0$ corresponds to the maximum of the function.

We could also have composed Taylor series around $x=0$ to obtain $$f(x)=\frac{1}{3}+\frac{8}{81} \left(\sqrt{3}-\pi \right) \pi x^2+\frac{8 \pi \left(\pi \left(27+2 \sqrt{3} \pi -3 \pi ^2\right)-9 \sqrt{3}\right) }{2187}x^4+O\left(x^6\right)$$ which a perfect approximation of the function for $-\frac 12 \leq x \leq \frac 12$ (maximum error $< 0.00012$ at the bounds).