TITLE: Prove : $\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$
QUESTION [13 upvotes]: If we assume that: $0\le x_1,x_2,\ldots,x_{10}\le\frac{\pi}{2} $ such that:
$$\sin^2 (x_1) +\sin^2 (x_2)+\cdots+\sin^2(x_{10})=1$$
 How to prove that:
$$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$$

REPLY [6 votes]: By Cauchy-Schwarz
$$\left(\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})\right)^2 \leq 9 \left( \sin^2(x_1) +\sin^2(x_2) +\cdots+\sin^2(x_{10})\right) =9$$
Thus
\begin{equation}
\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10}) \leq 3  \tag1
\end{equation}
Also, since $$0 \leq \cos(x_i) \leq 1$$ we 
$$\cos^2(x_i) \leq \cos(x_i)$$
Thus
\begin{equation}
\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10}) \geq \cos^2(x_1) +\cos^2(x_2) +\cdots+\cos^2(x_{10})=9 \tag2
\end{equation}
Combining (1) and (2) you get the desired result. 
P.S. How can I label equations?