<s>[INST] 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:
$$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$$  
 [/INST] 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
$$
\frac{\cos(x_1)+\cos(x_2)+\dots+\cos(x_{10})}{\sin(x_1)+\sin(x_2)+\dots+\sin(x_{10})}\ge3\tag{7}
$$
</s>