# A Trigonometric Inequality with Ordered Triple of Variables

### Solution 1

Since $\displaystyle \frac{\sin t}{t}$ is a decreasing function on $\displaystyle \left(0, \frac{\pi}{2}\right),$ we find that

$x \sin y < y \sin x, y \sin z < z \sin y, x \sin z < z \sin x.$

Therefore,

\begin{align} &(y + z) \sin x - (x - z) \sin y - (x + y) \sin z\\ &\qquad= (y \sin x - x \sin y) + (z \sin y - y \sin z) + (z \sin x - x \sin z) \gt 0. \end{align}

We can show that $\displaystyle \frac{\sin t}{t}$ is a decreasing function on $\displaystyle \left(0, \frac{\pi}{2}\right)$ without invoking derivatives:

\displaystyle \begin{align} \frac{\sin y}{y}-\frac{\sin x}{x}&=\frac{x \sin y-y\sin x}{xy}\\ &=\frac{x\sin y-x\sin x-y\sin x+x\sin x}{xy}\\ &=\frac{x(\sin y-\sin x)-(y-x)\sin x}{xy}\\ &=\frac{y-x}{xy}\Bigr[x\frac{\sin y-\sin x}{x-y}-\sin x\Bigr]\\ &=\frac{y-x}{xy}\Bigr[2x\frac{\sin \frac{y-x}{2}\cos \frac{y+x}{2}}{y-x}-\sin x\Bigr]\\ &\lt\frac{y-x}{xy}\Bigr[x\cos \frac{y+x}{2}-\sin x\Bigr]\\ &\lt\frac{y-x}{xy}(x\cos x-\sin x)=\frac{y-x}{xy}\cos x(x-\tan x)\\ &\lt 0 \end{align}

because, $y\gt x$ and $x\lt \tan x$.

$\displaystyle \frac{\sin y}{y}-\frac{\sin x}{x}\lt 0\Rightarrow x \sin y\lt y\sin x.$

### Solution 2

If $z\gt x,$ $x-z\lt 0.$ If $0\lt y\lt\displaystyle\frac{\pi}{2},$ $\sin y\gt 0.$ Since $(x-z)$ is negative and $\sin y$ is positive, $(x-z)\sin y$ is negative. Therefore, $(x+y)\sin x\gt (x+y)\sin x+(x-z)\sin y.$ Since $z\gt x,$ $(y+z)\gt (x+y)$ and $(y+z)\sin x\gt (x+y)\sin x.$ Therefore, $(x+y)\sin x+(x-z)\sin y \lt (y+z)\sin x.$

### Solution 3

The function $\displaystyle \frac{\sin t}{t}$ function is decreasing over $(0,\pi/2)$. Thus,

\displaystyle \begin{align} &\frac{\sin x}{x}\gt \frac{\sin y}{y}~\Rightarrow~y\sin x\gt x\sin y, \\ &\frac{\sin x}{x}\gt \frac{\sin z}{z}~\Rightarrow~z\sin x\gt x\sin z, \\ &\frac{\sin y}{y}\gt \frac{\sin z}{z}~\Rightarrow~z\sin y\gt y\sin z. \end{align}

$y\sin x+z\sin x+ z\sin y \gt x\sin y +x\sin z+y\sin z.$

Rearranging

$(y+z)\sin x \gt (x+y)\sin z+(x-z)\sin y.$

### Acknowledgment

I am grateful to Dan Sitaru for mailing me this problem, along with a solution of his (Solution 1). The problem has been published in the Crux Mathematicorum (VOLUME 43, NO. 9 November/Novembre 2017) as #4188.

Solution 2 is by DeplorableTwit; Solution 3 is by Amit Itagi.