Let $x,y>0$ then prove that:
$$|\cosh(x)-\cosh(y)|\geq|x-y|\sqrt{\sinh(x)\sinh(y)}.$$
I have tried to show it in a simple case when $y=x+1$ to delete terms $x-y$ then I don't differentiate because we can substitute $X=e^x$. Now we get a quartic and multiplying by $X^2$ and putting $Y=X^2$ we obtain a quadratic. Conclude is smooth now!
For the general case we can think to make $x-y$ constant but unfortunately we cannot do the same things as above.
How to show it? And then how to show it without derivative if possible?
Thanks you.
Edit :
As user Gary make a nice answer I complete it to show it without differentiation :
We want to show $x,y>0$ :
$$\sinh(x)\sinh(y)\leq \sinh^2\left(\frac{x+y}{2}\right)$$
Using $X=e^x$ and $Y=e^y$ the inequality after some algebraic manipulation is :
$$\frac{X}{Y}+\frac{Y}{X}-2\geq 0$$
See here to convince you .
Best Answer
\begin{align*} \left| {\cosh x - \cosh y} \right| & = 2\left| {\sinh \left( {\frac{{x - y}}{2}} \right)\sinh \left( {\frac{{x + y}}{2}} \right)} \right| \\ & \ge 2\frac{{\left| {x - y} \right|}}{2}\left| {\sinh \left( {\frac{{x + y}}{2}} \right)} \right| = \left| {x - y} \right|\left| {\sinh \left( {\frac{{x + y}}{2}} \right)} \right| \\& \ge \left| {x - y} \right|\sqrt {\sinh x\sinh y} \end{align*} since $|w|\leq|\sinh w|$ and $\sinh$ is log-concave on the positive axis.