Another Olympiad question:
Let $x, y, z\in\mathbb{R}^+$ satisfy $\frac{1}{\cosh x}+\frac{1}{\cosh y}+\frac{1}{\cosh z}=1$. Show that
$$
\sinh x \cdot \sinh y \cdot \sinh z \geq 16 \sqrt{2} .
$$
When does equality hold?
First I used $\sinh x\leq \cosh x$ to get
$$
1=\frac{1}{\cosh x}+\frac{1}{\cosh y}+\frac{1}{\cosh z}\leq \frac{1}{\sinh x}+\frac{1}{\sinh y}+\frac{1}{\sinh z}
$$
so
$$
1\leq \frac{1}{\sinh x}+\frac{1}{\sinh y}+\frac{1}{\sinh z}
$$
and then
$$
\sinh x \cdot \sinh y \cdot \sinh z \leq \sinh x \cdot \sinh y + \sinh x \cdot \sinh z + \sinh y \cdot \sinh z.
$$
But this is useless. I also thought of using
$$
\sinh x \cdot \sinh y \cdot \sinh z=\frac{1}{8}\left(e^x-e^{-x}\right)\left(e^y-e^{-y}\right)\left(e^z-e^{-z}\right)
$$
but this leads nowhere.
Any hints or ideas?
Best Answer
Let $$a=\frac1{\cosh x}, \qquad b=\frac1{\cosh y}, \qquad c=\frac1{\cosh z},$$ then $a, b, c>0$ and $a+b+c=1$. Since $\sinh x, \sinh y, \sinh z>0$, the desired inequality is equivalent to $$\sinh^2x\cdot\sinh^2y\cdot\sinh^2z\geq512,$$ which is further equivalent to $$(1-a^2)(1-b^2)(1-c^2)\geq 512a^2b^2c^2,\tag{$*$}$$ by using the elementary relation $\sinh^2x=\cosh^2x-1$.
Hint for proving $(*)$:
Full proof of $(*)$: