Prove the inequality $1 - \tanh(xy) \leq \cosh(x)^{-y}$

Question

Using some tricks in statistical mechanics I came across the inequality. $$ 1 - \tanh(xy) \leq \cosh(x)^{-y} $$ for all $x,y >0$. Do you have a proof (or counterexample)?

Answer 1

Let $\alpha$ be a positive real number greater than or equal to $1$.

Since $f_\alpha(z)=z^\alpha:\left[0,+\infty\right[\to\mathbb{R}$ is a convex function, it results that

$f_\alpha\left(\frac{z_1+z_2}{2}\right)\le\frac{f_\alpha(z_1)+f_\alpha(z_2)}{2}$ for any $z_1,z_2\in\left[0,+\infty\right[.$

Let $x,y$ be any two positive real numbers with $y\ge1$, we get that

$f_y\left(\frac{e^{x}+e^{-x}}{2}\right)\le\frac{f_y(e^{x})+f_y(e^{-x})}{2}\;\;$ that is

$\left(\frac{e^{x}+e^{-x}}{2}\right)^y\le\frac{e^{xy}+e^{-xy}}{2}$

$\frac{2}{e^{xy}+e^{-xy}}\le\left(\frac{e^{x}+e^{-x}}{2}\right)^{-y}$

$1-\frac{e^{xy}-e^{-xy}}{e^{xy}+e^{-xy}}=\frac{2e^{-xy}}{e^{xy}+e^{-xy}}<\frac{2}{e^{xy}+e^{-xy}}\le\left(\frac{e^{x}+e^{-x}}{2}\right)^{-y}$

$1-\tanh(xy)=1-\frac{e^{xy}-e^{-xy}}{e^{xy}+e^{-xy}}<\left(\frac{e^{x}+e^{-x}}{2}\right)^{-y}=\left(\cosh x\right)^{-y}.$

Hence,

$1-\tanh(xy)<\left(\cosh x\right)^{-y}\;$ for any $x,y\in\mathbb{R}^+$ with $y\ge1$.

Answer 2

It is possible to prove the inequality for all $x,y\in\mathbb{R}^+$.

Let $x$ and $y$ be any two positive real numbers.

Since $$\frac{2}{e^{xy}+e^{-xy}}<\frac{2+\left(e^\frac{xy}{2}-e^\frac{-xy}{2}\right)^2} {e^{xy}+e^{-xy}}=1$$ we get that

$$\frac{2e^{-xy}}{e^{xy}+e^{-xy}}<e^{-xy}.\;\;\;\;\;(*)$$

Since $$e^{-x}=\frac{2}{2e^x}<\frac{2}{e^x+e^{-x}}$$ it follow that $$e^{-xy}<\left(\frac{2}{e^x+e^{-x}}\right)^y.\;\;\;\;\;(**)$$

From $(*)$ and $(**)$ we get that $$\frac{2e^{-xy}}{e^{xy}+e^{-xy}}<\left(\frac{2}{e^x+e^{-x}}\right)^y.$$

Moreover, $$1-\tanh(xy)=1-\frac{e^{xy}-e^{-xy}}{e^{xy}+e^{-xy}}=\frac{2e^{-xy}}{e^{xy}+e^{-xy}}<\left(\frac{2}{e^x+e^{-x}}\right)^y=\left(\cosh x\right)^{-y}.$$

Hence,

$$1-\tanh(xy)<\left(\cosh x\right)^{-y}\;\;\;\text{for any}\;\;x,y\in\mathbb{R}^+.$$

Answer 3

Let $u = xy$. We have to prove that $1-\tanh u \le (\cosh x)^{-u/x}$.

Since $(\cosh x)^{-u/x} = \mathrm{e}^{-\frac{u}{x}\ln \cosh x} = \mathrm{e}^{-\frac{u}{x}\ln \frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}} \ge \mathrm{e}^{-\frac{u}{x}\ln \mathrm{e}^x} = \mathrm{e}^{-u}$, it suffices to prove that $1-\tanh u \le \mathrm{e}^{-u}$ or $1 - \frac{\mathrm{e}^u - \mathrm{e}^{-u}}{\mathrm{e}^u + \mathrm{e}^{-u}} \le \mathrm{e}^{-u}$ or $\mathrm{e}^u + \mathrm{e}^{-u} \ge 2$ which is true. We are done.