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:

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


3 Answers 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 $\arctan(1/3)$:

$\hspace{2cm}$crossing image

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

  • $\begingroup$ But is ($4$) necessarily true ? i mean when the angle 45 degree or 30 degree. $\endgroup$
    – Frank
    Aug 10, 2012 at 22:38
  • $\begingroup$ @MohammedAl-mubark: It is true for all $x\in[0,\frac\pi2]$. Look at the diagram: $\cos(x)-3\,\sin(x)$ is negative precisely when $\cos(x)+3\,\sin(x)-\frac6{\sqrt{10}}$ is positive, and vice-versa. $\endgroup$
    – robjohn
    Aug 10, 2012 at 23:32

By Cauchy-Schwarz

$$\left(\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})\right)^2 \leq 9 \left( \sin^2(x_1) +\sin^2(x_2) +\cdots+\sin^2(x_{10})\right) =9$$


\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

$$\cos^2(x_i) \leq \cos(x_i)$$


\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?

  • $\begingroup$ Af for your P.S., try with the tag \tag{put_here_a_label} at the end of your equations. $\endgroup$ Aug 10, 2012 at 15:50
  • 1
    $\begingroup$ Shouldn't it be $(\sin(x_1)+...+\sin(x_{10}))^2\le10(\sin^2(x_1)+...+\sin^2(x_{10}))$ from Cauchy-Schwarz? $\endgroup$ Oct 18, 2012 at 0:18

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.


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