Prove : $\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$ 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:
$$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$$  
 A: 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?
A: Transform to $y_i=\sin^2 x_i$. Then $\sum_iy_i=1$, and with $\sin x_i=\sqrt{y_i}$ and $\cos x_i=\sqrt{1-y_i}$ the inequality becomes
$$
\sum_i\left(\sqrt{1-y_i}-3\sqrt{y_i}\right)\ge0\;.
$$
The left-hand side is $10$ times the average value of $f(y)=\sqrt{1-y}-3\sqrt y$ at the $y_i$. The graph of $f$ has an inflection point at $y=1/\left(1+3^{-2/3}\right)\approx0.675$ and is convex to its left. Either none or one of the $y_i$ can be to its right. If none are, the average value of $f$ is greater or equal to the value at the average, $f(1/10)=0$. If one is, say, $y_i$, then we can bound the average value of the remaining ones by the value at their average, so in this case
$$
\sum_i\left(\sqrt{1-y_i}-3\sqrt{y_i}\right)\ge\sqrt{1-y_1}-3\sqrt{y_1}+9\sqrt{1-(1-y_1)/9}-27\sqrt{(1-y_1)/9}\;.
$$
This is non-negative with a single root at $1/10$, so the inequality holds in both cases.
A: Since
$$
\sin^2(x_1)+\sin^2(x_2)+\dots+\sin^2(x_{10})=1\tag{1}
$$
we have
$$
\cos^2(x_1)+\cos^2(x_2)+\dots+\cos^2(x_{10})=9\tag{2}
$$
Therefore,
$$
\frac{\cos^2(x_1)+\cos^2(x_2)+\dots+\cos^2(x_{10})}{\sin^2(x_1)+\sin^2(x_2)+\dots+\sin^2(x_{10})}=9\tag{3}
$$
Note that for $0\le x\le\frac\pi2$, we have that
$$
\cos(x)-3\sin(x)\ge0\quad\Leftrightarrow\quad\cos(x)+3\sin(x)-\frac6{\sqrt{10}}\le0\tag{4}
$$
since they both change signs only at $\arctan(1/3)$:
$\hspace{2cm}$
Therefore, $(4)$ immediately gives
$$
\cos^2(x)-9\,\sin^2(x)\le\frac6{\sqrt{10}}(\cos(x)-3\sin(x))\tag{5}
$$
Summing $(5)$ and applying $(3)$, we get
$$
\begin{align}
0
&=\sum_{i=1}^{10}\cos^2(x_i)-9\,\sin^2(x_i)\\
&\le\frac6{\sqrt{10}}\sum_{i=1}^{10}\cos(x_i)-3\sin(x_i)\tag{6}
\end{align}
$$
Inequality $(6)$ yields
$$
\frac{\cos(x_1)+\cos(x_2)+\dots+\cos(x_{10})}{\sin(x_1)+\sin(x_2)+\dots+\sin(x_{10})}\ge3\tag{7}
$$
