TITLE: $ \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}. $
QUESTION [5 upvotes]: Suppose that $x_1,.\dots x_{10}\in[0,\frac{\pi}{2}]$ and that
  $$
\sin^2x_1+\dots \sin^2x_{10}=1.
$$
  Prove that
  $$
3(\sin x_1+\dots \sin x_{10})\leq \cos x_1 +\dots +\cos x_{10}.
$$

REPLY [2 votes]: 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!