Establishing a bound of the form $\frac{\cosh(x)}{\cosh(y)} \le \cosh(M(x-y)) $

Question

Suppose we have $x, y \in \mathbb{R}$ where $0 \le y \le x$. I'm trying to establish a bound of the form $$ \frac{\cosh(x)}{\cosh(y)} \le \cosh(M(x-y)) $$ where $M$ is a positive constant.

Trial and error has shown that this probably holds for $M = 2$, but I'm having a hard time proving this.

Does anyone have any idea how I could establish this?

Answer 1

It is not true. At $(x,y)=(7.4,7.2)$ we calculate numerically: \begin{align*} \frac{\cosh(x)}{\cosh(y)}=\frac{\cosh(7.4)}{\cosh(7.2)}&\approx 1.221\\ \cosh(2(x-y))=\cosh(0.4)&\approx 1.081 \end{align*}

In fact, there can be no $M$ that works:

Expanding LHS at small $x-y$ gives $$ \frac{\cosh(x)}{\cosh(y)}=1+\tanh(y)(x-y)+o(x-y) $$ but the RHS expands as $$ \cosh(M(x-y))=1+o(x-y) $$ so as long as $y$ is large, you can find $x-y$ small enough to beat the bound. The example suggests something along the line $x,y\approx e^M$ and $x-y\approx e^{-M}$.

Answer 2

The estimate does not hold for any positive constant $M$ and all $0 < y \le x$:

$$ \cosh(x) \le \cosh(y) \cdot \cosh(M(x-y)) $$ is equivalent to $$ \frac{\cosh(x)-\cosh(y)}{x-y} \le \cosh(y) \cdot \frac{\cosh(M(x-y))-1}{M(x-y)} \cdot M $$ and taking the limit $x \to y$ gives $$ \sinh(y) \le \cosh(y) \cdot 0 \cdot M = 0 $$ which is impossible for positive $y$.