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:
$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$
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:
$\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$
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 $\arctan(1/3)$:
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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} $
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?
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.