通过级数求加权平均不等式的最佳系数

1+1=？

\[
\begin{aligned}
& \forall x \in\left[0, \frac{\pi}{2}\right) \\
& \left(\frac{\pi^2}{6}-1\right) \frac{x}{1-\frac{4}{\pi^2 x^2}}+\left(2-\frac{\pi^2}{6}\right) \frac{x}{\sqrt{1-\frac{4}{\pi^2} x^2}} \leq \tan x
\end{aligned}
\]
该不等式还是比较强的，请问(π^2/6-1)这个最佳系数是如何求出的？
在QQ群中有人说可以通过级数求出，请问有人知道这一方法吗？

战巡

不妨设
\[a\frac{x}{1-\frac{4}{\pi^2}x^2}+b\frac{x}{\sqrt{1-\frac{4}{\pi^2}x^2}}\approx \tan(x)\]
其中$a>0,b>0$

那么
\[a+b\sqrt{1-\frac{4}{\pi^2}x^2}\approx(1-\frac{4}{\pi^2}x^2)\frac{\tan(x)}{x}\]
\[b^2(1-\frac{4}{\pi^2}x^2)\approx\left[(1-\frac{4}{\pi^2}x^2)\frac{\tan(x)}{x}-a\right]^2\]

右边泰勒展开，得到
\[\left[(1-\frac{4}{\pi^2}x^2)\frac{\tan(x)}{x}-a\right]^2=(a-1)^2+2(a-1)(\frac{4}{\pi^2}-\frac{1}{3})x^2+\left[2(a-1)(\frac{4}{3\pi^2}-\frac{2}{15})+(\frac{4}{\pi^2}-\frac{1}{3})^2\right]x^4+o(x^4)\]
比对系数，得到
\[\begin{cases}(a-1)^2=b^2\\2(a-1)(\frac{4}{\pi^2}-\frac{1}{3})=-\frac{4b^2}{\pi^2}\end{cases}\]
\[\begin{cases}a=\frac{\pi^2}{6}-1\\b=2-\frac{\pi^2}{6}\end{cases}\]

Aluminiumor

待定系数
$$a\cdot\frac{x}{1-4x^2/\pi^2}+b\cdot\frac{x}{\sqrt{1-4x^2/\pi^2}}\sim\tan x$$
易知
$$\frac{x}{1-4x^2/\pi^2}\sim x+\frac{4}{\pi^2}x^3+O\left(x^4\right)$$
$$\frac{x}{\sqrt{1-4x^2/\pi^2}}\sim x+\frac{2 }{\pi ^2}x^3+O\left(x^4\right)$$
$$\tan x\sim x+\frac{x^3}{3}+O\left(x^4\right)$$
令 $$a+b=1$$
$$\frac{4}{\pi^2}a+\frac{2}{\pi^2}b=\frac13$$
解之即得
$$a=\frac{\pi^2}{6}-1,b=2-\frac{\pi^2}{6}$$