Prove that $(n-1) \sum_1^n \cot(\theta_i) \leq \sum_1^n \tan(\theta_i) $ 
n is a positive integer and $\theta_i$ is such that $ 0^\circ \leq \theta_i \leq 90^\circ $ for all positive integers $i \leq n$ and $\sum_1^n \cos^2(\theta_i) = 1$. Prove that $(n-1) \sum_1^n \frac{1}{\tan(\theta_i)} \leq \sum_1^n \tan(\theta_i) $

I tried this question numerous times, but always arrived in a dead end, may someone please help?
 A: (So, let's break down how one can approach this problem. Bearing in mind that it most probably has a reasonable answer, which is why you would have come across this problem.) The $n-1$ term is often not nice to deal with, and one of the things that it suggests is that we perform an inequality with only $n-1$ terms, instead of $n$ terms.
Let $\cos \theta_i = x_i \geq 0 $. We have $ \sin ^2 \theta = 1 - \cos^2 \theta = \sum _{j\neq i } x_j ^2 $. This suggests a natural $n-1$ term inequality to consider.   
Recall from Power Mean inequality that 
$$ \frac{n-1} { \sum_{j\neq i } \frac{1}{x_i} } \leq  \frac{\sum_{j\neq i} x_j} { n-1} \leq \sqrt{ \frac{ \sum _{j\neq i } x_j^2 } { n-1} }.$$
As such, $\frac{ \sin \theta_i}{\cos \theta_i } = \frac{ \sqrt{ \sum _{j\neq i} x_j ^2 }} { x_i} \geq \frac{1}{ \sqrt{n-1} } \frac { \sum_{j\neq i} x_j} {x_i}$.
Also, $\frac{\cos \theta_i} { \sin \theta_i} = \frac{x_i} { \sqrt{ \sum_{j\neq i } x_j^2 }}  \leq \frac{x_i}{ (n-1) \sqrt{n-1} } \sum_{j\neq i } \frac{1}{x_j}$.
Hence, we have 
$$ \sum \tan \theta_i \geq \frac{1}{\sqrt{n-1}} \sum_{j\neq i} \frac{x_j}{x_i} \geq (n-1) \sum \cot \theta_i.$$

[This approach is wrong]
(So, let's break down how one can approach this problem. Bearing in mind that it most probably has a reasonable answer, which is why you would have come across this problem.) The $n-1$ term is often not nice to deal with. We would very much prefer for it to be $n$, but multiplying it by $\frac{n}{n-1}$ does not resolve this issue. Instead (most of the time), we should add $ \sum \frac{1}{ \tan \theta_i }$ to both sides and hope that things work out.
Hint: Prove that
$$ n \sum  \frac{\cos \theta_i } { \sin \theta_i} \leq \sum \frac{1}{ \sin \theta_i \cos \theta_i}, $$
using the rearrangement inequality.
Obvious reminder: Use $1 = \sum \cos^2 \theta $.
So, I was originally thinking of applying the Chebyshev inequality to get that
$$ n \sum  \frac{\cos \theta_i } { \sin \theta_i} \leq \sum \frac{1}{ \sin \theta_i \cos \theta_i} \sum \cos^2 \theta_i. $$
However, the two sequences are not oppositely sorted, which is why this approach is wrong.
