Prove that
$$\sinh(\cosh(x)) \geq \cosh(\sinh(x))$$
I tried to tackle this problem by integrating both lhs and rhs, in order to get two functions who show clearly that inequality holds.
I've struggled for this problem a little bit, i don't know if there's any trick that can help. Maybe knowing that
$$\cosh^{-1}(x) = \pm \ln\left(x + \sqrt{x^2 – 1}\right)$$
Can help?
Best Answer
Here's a solution with very little calculus. First, an identity: \begin{align*} \sinh^2(a+b) - \sinh^2(a-b) &= (\sinh a\cosh b + \cosh a\sinh b)^2 - (\sinh a\cosh b - \cosh a\sinh b)^2 \\ &= 4\sinh a \cosh a\sinh b \cosh b \\ &= \sinh(2a)\sinh(2b) \end{align*} Taking $a=e^x/2$ and $b=e^{-x}/2$, we get \begin{align*} \sinh^2(\cosh x) - \sinh^2(\sinh x) &= \sinh(e^x) \sinh(e^{-x}) \\ &\ge e^xe^{-x} &&\text{(since $\sinh t\ge t$ for $t\ge 0$)} \\ &= 1 \\ &= \cosh^2(\sinh x) - \sinh^2(\sinh x) \end{align*} Cancelling $\sinh^2(\sinh x)$ and taking square roots gives the desired inequality.
Calculus is needed here only to justify the inequality $\sinh t\ge t$ (for $t\ge 0$).
Update: Another nice thing about this method is that it points the way to a more exact inequality. It turns out that $$ \sinh u\sinh v \ge \sinh^2\sqrt{uv} $$ for $u,v\ge 0$. (Proof 1: $\frac12(u^{2m+1}v^{2n+1} + v^{2m+1}u^{2n+1})\ge (uv)^{m+n+1}$ by AM/GM; divide by $(2m+1)!\,(2n+1)!$ and apply $\sum_{m=0}^\infty \sum_{n=0}^\infty$. Proof 2: Check that $t\mapsto\ln\sinh(e^t)$ is convex (for all $t$) by computing its second derivative.)
Applying this with $u=e^x$ and $v=e^{-x}$ above, we get $$ \sinh^2(\cosh x) \ge \cosh^2(\sinh x) + \underbrace{\sinh^2(1) - 1}_{\approx 0.3811} $$ with equality when $x=0$.