Proving $\cosh^2(\cosh x) - \sinh^2 (\sinh x) \geq2$ for real $x$

Question

Prove that $$\cosh^2(\cosh x) - \sinh^2 (\sinh x) \geq2, \qquad\forall x \in\Bbb{R}$$

It is hard to derive inequality from hyperbolic functions. I have tried hyperbolic expansions, addition theorem, and the definition of the functions.

Please help thank you!

Answer 1

1. Using the trigonometric identities

$$ \cosh^2 A - \sinh^2 A = 1 \qquad \text{and}\qquad \cosh^2 A - \cosh^2 B = \sinh A \sinh B, $$

we find that

\begin{align*} &\cosh^2(\cosh x) - \sinh^2(\sinh x) \\ &= \cosh^2(\cosh x) - \cosh^2(\sinh x) + 1 \\ &= \sinh(e^x)\sinh(e^{-x}) + 1. \end{align*}

So the desired inequality is equivalent to

$$ \sinh(e^x)\sinh(e^{-x}) \geq 1, \qquad \forall x \in \mathbb{R}. \tag{1} $$

2. Now we will show that $\text{(1)}$ is indeed true. Writing $f(x) = \log (\sinh(e^x)\sinh(e^{-x}))$ and computing its derivative, we get

$$ f'(x) = e^x \coth(e^x) - e^{-x}\coth(e^{-x}). $$

Then by noting that

$$ \frac{\mathrm{d}}{\mathrm{d}t} \, t \coth t = \frac{\cosh t \sinh t - t}{\sinh^2 t} > 0, \qquad \forall t > 0, $$

it follows that $t \mapsto t \coth t$ is strictly increasing on $t > 0$. From this, we find that $f'(x)$ is also strictly increasing in $x$ and $f'(0) = 0$ holds. Altogether, we know that $x = 0$ is the global minimum point for $f(x)$, hence proving

$$ \sinh(e^x)\sinh(e^{-x}) = e^{f(x)} \geq e^{f(0)} = \sinh^2(1) > 1 $$

as desired.

Answer 2

Consider that you look for the minimum value of function $$f(x)=\cosh^2(\cosh (x)) - \sinh^2 (\sinh (x)) $$ COmpute the first derivative and simplify it $$f'(x)=\sinh (x) \sinh (2 \cosh (x))-\sinh (2 \sinh (x)) \cosh (x)$$ The trivial solution $x=0$ is clear because of the first and third hyperbolic sines.

So $f(0)=\cosh^2(\cosh (0))\sim 2.38 > 2$

Now, $$f''(x)=\cosh (x) (\sinh (2 \cosh (x))-2 \cosh (x) \cosh (2 \sinh (x)))+$$ $$\sinh (x) (2 \sinh (x) \cosh (2 \cosh (x))-\sinh (2 \sinh (x)))$$ $$f''(0)=\sinh (2)-2 \sim 1.63 > 0$$ So, $x=0$ is $\color{red}{\large a}$ minimum. However, as @Martin R commented, I did not prove that this is the global minimum.