Showing $\sum_{i=1}^n\tan\alpha_i\geq (n-1)\cdot \sum_{i=1}^n\cot\alpha_i$, for real $\alpha_i\in(0,\pi/2)$ with $\sum_{i=1}^n\cos^2\alpha_i=1$

Question

Real numbers $\alpha_1,\ldots,\alpha_n \in \left(0,\ \frac{\pi} 2\right)$ satisfy the condition $\sum_{i=1}^n\cos^2\alpha_i=1$. Prove that $$\sum_{i=1}^n\tan\alpha_i\geq (n-1)\cdot \sum_{i=1}^n\cot\alpha_i$$

This is from the final part of the book, where no technique is suggested.

I tried using the observation that $n-1 = \sum\sin^2\alpha_i$, but substituting this into the inequality we obtain the inequality that is false in general (without the assumption $\sum\cos^2\alpha_i=1$).

My most promising idea is the following. Thanspose the condition into: $$n\cdot \sum_{i=1}^n\cot\alpha_i \leq \sum_{i=1}^n\left(\tan\alpha_i + \cot\alpha_i\right).$$ $$n\cdot \sum_{i=1}^n\cot\alpha_i \leq \sum_{i=1}^n\frac 1{\sin\alpha_i\cdot \cos\alpha_i}.$$ $$n\cdot \sum_{i=1}^n\frac {\cos^2\alpha_i}{\sin\alpha_i\cdot \cos\alpha_i} \leq \sum_{i=1}^n\frac 1{\sin\alpha_i\cdot \cos\alpha_i}.$$ Then we could use Chebyshev's inequality for anti-monotone sequences (I don't know the name in English). Assuming $\alpha_i\leq \alpha_{i+1}$, the sequence $a_i = \cos^2\alpha_i$ is decreasing and $b_i = 1/(\sin\alpha_i\cos\alpha_i)$ is increasing provided $\alpha_i\geq \pi/4$. Therefore we have the proof for such angles: $$n\cdot \sum_{i=1}^n\frac {\cos^2\alpha_i}{\sin\alpha_i\cdot \cos\alpha_i} \stackrel{(\star)}\leq \left(\sum_{i=1}^n \cos^2\alpha_i\right)\cdot \sum_{i=1}^n\frac 1{\sin\alpha_i\cdot \cos\alpha_i} = \sum_{i=1}^n\frac 1{\sin\alpha_i\cdot \cos\alpha_i}.$$ Unfortunately, one angle $\alpha_1$ can be smaller than $\pi/4$ and the proof doesn't go. What's more, the inequality (*) also doesn't hold in general (if one angle is smaller than $\pi/4$).

I started thinking if the conclusion is false, but I couldn't find the counterexample (I used spreadsheet).

Answer 1

Let $\tan^2\alpha_i=x_i$.

Thus, the condition gives $$\sum_{i=1}^n\frac{1}{1+x_i}=1$$ and we need to prove that: $$\sum_{i=1}^n\sqrt{x_i}\geq(n-1)\sum_{i=1}^n\frac{1}{\sqrt{x_i}}.$$ Now, see here: Prove that $\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}+\cdots+\frac{1}{\sqrt{x_n}} \right )$