Show that $\sin(x)\sin(y)\sin(z) \leq \frac{1}{8}$

Question

Let $(x,y,z) \in (\mathbb{R}^+)^3$ such that $x + y + z \leq \frac{\pi}{2}$.

Show that $$\sin(x)\sin(y)\sin(z) \leq \frac{1}{8}$$

I have a solution using convexity of $\sin$ but I am looking for a method wich does not use convexity.

EDIT: I don't want to use convexity because I'd like to do the exercise in accordance with the program of high school (although I do not know if this is possible)

Answer 1

first we prove when $x+y+z=\dfrac{\pi}{2},\sin(x)\sin(y)\sin(z) \leq \dfrac{1}{8}$:

$\sin(y)\sin(z)=\dfrac{cos(y-z)-cos(y+z)}{2} \le \dfrac{1-cos(y+z)}{2}=\dfrac{1-sin(x)}{2}$

because $cos(y-z) \le 1 $ when $y=z$ get "= 1"

$\sin(x)\sin(y)\sin(z) \le \dfrac{\sin(x)-\sin^2(x)}{2}=\dfrac{-(\sin(x)-0.5)^2+0.25}{2} \le \dfrac{1}{8}$

when $\sin(x)=0.5$ get $\dfrac{1}{8} \implies x=y=z=\dfrac{\pi}{6}$

in case $x+y+z\le \dfrac{\pi}{2}$ ,let $k=\dfrac {x+y+z}{\dfrac{\pi}{2}}\le1,x=kx',y=ky'.z=kz' \implies x'+y'+z'= \dfrac{\pi}{2},$

in $(0,\dfrac{\pi}{2}],0<\sin(x) \le sin(x') \implies \sin(x)\sin(y)\sin(z) \le \sin(x')\sin(y')\sin(z')$

when $k=1$ get "="

Answer 2

Use the fact that for a continues function on a compact region the maximum can be in either a point where $f' = 0 $ or on the edge of the region.