Prove : $\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3$

Solution 1:

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 $$ \frac{\cos(x_1)+\cos(x_2)+\dots+\cos(x_{10})}{\sin(x_1)+\sin(x_2)+\dots+\sin(x_{10})}\ge3\tag{7} $$

Solution 2:

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?