$ \sin^2x_1+\dots \sin^2x_{10}=1$ implies $ 3(\sin x_1+\dots \sin x_{10})\leq \cos x_1 +\dots +\cos x_{10}. $ 
Suppose that $x_1,.\dots x_{10}\in[0,\frac{\pi}{2}]$ and that
  $$
\sin^2x_1+\dots \sin^2x_{10}=1.
$$
  Prove that
  $$
3(\sin x_1+\dots \sin x_{10})\leq \cos x_1 +\dots +\cos x_{10}.
$$

 A: First of all notice that $$n\sum_{i=1}^ny_i^2 \ge \left( \sum_{i=1}^n y_i \right)^2$$Proof: Notice that $$\sum_{i=1}^{n-1} \sum_{j=i+1}^n (y_j - y_i)^2 = (n-1)\left( \sum_{i=1}^ny_i^2 \right) - 2\left(\sum_{i=1}^{n-1}\sum_{j=i+1}^n y_iy_j\right) \ge 0$$It follows that 
$$n\left( \sum_{i=1}^ny_i^2 \right) - 2\left(\sum_{i=1}^n y_i^2 + \sum_{i=1}^{n-1}\sum_{j=i+1}^n y_iy_j\right)$$ $$=n\left( \sum_{i=1}^ny_i^2 \right) - \left(\sum_{i=1}^ny_i\right)^2 \ge 0$$ proving the inequality. 
Now notice that $$\cos(x_1) = \sqrt{\sum_{i=1, i\ne 1}^{10}\sin^2(x_i)},$$ $$ \cos(x_2) = \sqrt{\sum_{i=1, i\ne 2}^{10}\sin^2(x_i)}$$ $$...$$. Now observe that $$9\sum_{i=1, i\ne 1}^{10}\sin^2(x_i) \ge \left( \sum_{i=1, i\ne 1}^{10} \sin(x_i) \right)^2$$ from the first inequality, implying that $$\sqrt{\sum_{i=1, i\ne 1}^{10}\sin^2(x_i)} \ge \frac{\left( \sum_{i=1, i\ne 1}^{10} \sin(x_i) \right)}3$$, i.e. $$\cos(x_1) \ge \frac{\left( \sum_{i=1, i\ne 1}^{10} \sin(x_i) \right)}3$$ Similarly $$\cos(x_2) \ge \frac{\left( \sum_{i=1, i\ne 2}^{10} \sin(x_i) \right)}3$$ $$...$$. Adding all these inequalities, we get $$\left(\cos(x_1) + \cos(x_2) + ... + \cos(x_{10})\right) \ge \frac{\left( \sum_{i=1, i\ne 1}^{10} \sin(x_i) \right)}3 + \frac{\left( \sum_{i=1, i\ne 2}^{10} \sin(x_i) \right)}3 + ... + \frac{\left( \sum_{i=1, i\ne 10}^{10} \sin(x_i) \right)}3$$ $$=3\left(\sin(x_1) + \sin(x_2) + ... + \sin(x_{10})\right)$$ proving the inequality.
A: TL method helps here very well!!!
Let $\sin x_i=\sqrt{\frac{a_i}{10}}$, where $a_i\geq0$. 
Hence,  $\sum\limits_{i=1}^{10}a_i=10$ and we need to prove that $$\sum\limits_{i=1}^{10}\left(\cos x_i-3\sin x_i\right)\geq0$$ or
$$\sum_{i=1}^{10}\left(\sqrt{10-a_i}-3\sqrt{a_i}\right)\geq0$$ or
$$\sum_{i=1}^{10}\left(\frac{1-a_i}{\sqrt{10-a_i}+3\sqrt{a_i}}+\frac{a_i-1}{6}\right)\geq0$$ or
$$\sum_{i=1}^{10}\frac{(a_i-1)\left(\sqrt{10-a_i}+3\sqrt{a_i}-6\right)}{\sqrt{10-a_i}+3\sqrt{a_i}}\geq0$$ or
$$\sum_{i=1}^{10}\frac{(a_i-1)^2\left(\frac{3}{\sqrt{a_i}+1}-\frac{1}{3+\sqrt{10-a_i}}\right)}{\sqrt{10-a_i}+3\sqrt{a_i}}\geq0,$$ which is obvious.
Done!
