my professor gives me this result :

Let $0<x_i<\frac{\pi}{2}$ be $n$ real numbers then we have : $$\Big(\sum_{i=1}^{n}x_i\Big)\Big(\tan\Big(\frac{\sum_{i=1}^{n}x_i^2}{\sum_{i=1}^{n}x_i}\Big)\Big)\Big(\sum_{i=1}^{n}\tan(x_i)\Big)\geq n\Big(\sum_{i=1}^{n}x_i\tan(x_i)\Big)\tan\Big(\frac{\sum_{i=1}^{n}x_i}{n}\Big)$$

The case $n=1$ is an equality .

For the case $n=2$ i try to use the derivatives but it's very ugly .

I try to use the inequality with the conidtion above :

$$\tan(x)\geq x$$

But the inequality is too sharp .

I try also a reasoning with induction but I can't prove some steps.

So any hints would be very appreciated .

Thanks a lot for sharing your time and knowledge .


1 Answer 1


Proof for $n=2$:

WLOG, assume that $x\ge y$.

Since $z\to \tan z$ is convex on $(0, \frac{\pi}{2})$, from the first order characterization of differentiable convex functions, we have $$\tan s \ge \tan t + (1 + \tan^2 t)(s - t), \quad \forall s, t\in (0, \frac{\pi}{2}).$$ By letting $s = \frac{x^2+y^2}{x+y}$ and $t = \frac{x+y}{2}$, we have $$\tan \frac{x^2+y^2}{x+y} \ge \tan \frac{x+y}{2} + \big(1 + \tan^2 \frac{x+y}{2}\big)\frac{(x-y)^2}{2(x+y)}.$$ Thus, it suffices to prove that \begin{align} &(x + y)\Big[\tan \frac{x+y}{2} + \big(1 + \tan^2 \frac{x+y}{2}\big)\frac{(x-y)^2}{2(x+y)}\Big] (\tan x + \tan y)\\ \ge\ & 2(x\tan x + y\tan y)\tan \frac{x+y}{2}. \tag{1} \end{align} With the substitutions $x = u + v, \ y = u - v$ for $u > v \ge 0, \ u+v < \frac{\pi}{2}$, the inequality (1) is written as $$\frac{2v\tan u\,\, (1 + \tan^2 u)(2v - \sin 2v)}{\cos^2 v\,\, (1-\tan u \tan v)(1 + \tan u \tan v)}\ge 0$$ which is obvious.

Remark: This method may not work for $n> 2$.


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