# $\sin^2x_1+\dots \sin^2x_{10}=1$ implies $3(\sin x_1+\dots \sin x_{10})\leq \cos x_1 +\dots +\cos x_{10}.$ [duplicate]

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

## marked as duplicate by Daniel Fischer♦Jul 20 '17 at 16:07

• If $x_1 = x_2 = \ldots = x_{10} \in [0,\frac{\pi}{2}]$ such that $\sin^2 x_1 = \frac{1}{10} \implies \sin x_1 = \sqrt{\frac{1}{10}}$ and $\cos x_1 = \sqrt{\frac{9}{10}}$, then, $$\sum_{i=1}^{10} \cos x_i = 3 \left( \sum_{i=1}^{10} \sin x_i \right)$$ If $x_1 = \frac{\pi}{2}$ and $x_2 = x_3 = \ldots = x_{10} = 0$ then, $$\sum_{i=1}^{10} \cos x_i = 9 \geq 3 = 3 \left( \sum_{i=1}^{10} \sin x_i \right)$$ If we deviate from the first case of equal $x_i$'s (when equality is achieved), the RHS tends to become larger and larger than the LHS. I'll try to prove this. – Dhruv Kohli - expiTTp1z0 Jul 8 '17 at 18:25
• David, you're not going to learn how to solve contest math questions without trying yourself to do so. (I'm referring to your initial tag "Olympiad problems = contest math".) I see no thoughts from you on the question, no effort or attempt from you within your question body. We're not here to "train you" by giving you answers to questions you'll need to master. – Namaste Jul 8 '17 at 18:30
• What have you tried, David? What is the source of this question? Which Olymiad? – Namaste Jul 8 '17 at 19:00
• I apologize if I wasn't clear with my intentions. I spent a good 3 hours working on this problem, writing it in different forms using various identities and trying use the Cauchy-Schwarz inequality. I was aware of the extremal cases from the beginning (one always should). I've never posted a 'contest problem' before so was unaware of protocol. I'm seriously stumped but do not expect to be handed an answer by any means, just perhaps a tiny nudge. The problem was posed by Alexandru Lupas in 1987 in the American Mathematics Monthly; it was never an olympiad problem but I feel it has that flavour. – David Peter Biddlesworth Jul 8 '17 at 19:32
• @Macavity I don't think that it's duplicate exactly, because my solution of this problem is different than all other solutions. – Michael Rozenberg Jul 9 '17 at 5:59

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!

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.