If $\frac1{\cosh x}+\frac1{\cosh y}+\frac1{\cosh z}=1$, prove that $\sinh x \cdot \sinh y \cdot \sinh z \geq 16 \sqrt{2}$.

Question

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.

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Any hints or ideas?

Answer 1

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 $(*)$:

Note that $1-a^2=(b+c)(a+a+b+c)$, then use AM-GM.

Full proof of $(*)$:

By AM-GM, we have\begin{align*}1-a^2&=(a+b+c)^2-a^2=(b+c)(a+a+b+c)\\&\geq 2b^{1/2}c^{1/2}\cdot4a^{1/2}b^{1/4}c^{1/4}=8a^{1/2}b^{3/4}c^{3/4}.\end{align*}For the same reason we have $1-b^2\geq 8a^{3/4}b^{1/2}c^{3/4}, 1-c^2\geq 8a^{3/4}b^{3/4}c^{1/2}$, then $(*)$ clearly follows. Also, the equality holds for $a=b=c\Longleftrightarrow x=y=z$.