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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?

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  • $\begingroup$ Jensen is useful only when we have a function and a constant, not with composite functions... Even if I think there are some similarities... $\endgroup$ – james42 May 26 '15 at 18:11
  • $\begingroup$ cheers for teaching me something new :). $\endgroup$ – Chinny84 May 26 '15 at 18:12
  • $\begingroup$ Physics guy here. I don't have any idea how to do math proofs, but isn't it true that a "small function" acting on a "big function" is always larger than the alternative? Since cosh(x)>sinh(x) for all x, this is a no brainer, right? This is how I remember e^pi>pi^e. Anyway, I've never spoken on Math.SE and generally run in fear from mathematicians, so please, let me down gently! $\endgroup$ – user1717828 May 27 '15 at 4:03
  • $\begingroup$ @user1717828 $\ln x$ is smaller than $x^2$, but $\ln (x^2) = 2\ln x$ is smaller than $(\ln x)^2$. $\endgroup$ – user21467 May 27 '15 at 4:21
  • $\begingroup$ @StevenTaschuk, aaaaannnnnddd this is why our field would crumble without mathematicians keeping us in check. Thanks! $\endgroup$ – user1717828 May 27 '15 at 4:43
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For any $y \ge 0$, notice

$$e^y - 1 = \int_0^y e^x dx \ge \int_0^y (1+x) dx \ge \int_0^y \left(1+\frac{x} {\sqrt{1+x^2}}\right)dx = y + \sqrt{1+y^2} - 1$$ we have this little inequality: $$\sqrt{1+y^2} - y = \frac{1}{\sqrt{1+y^2} + y} \ge e^{-y}$$

Using MVT, we can find a $\xi \in (y,\sqrt{1+y^2})$ such that

$$\sinh\sqrt{1+y^2} - \sinh(y) = \cosh(\xi)\left(\sqrt{1+y^2} - y\right) \ge \cosh(\xi) e^{-y} \ge e^{-y}$$

Since $e^{-y} = \cosh(y) - \sinh(y)$, this leads to $$ \sinh\sqrt{1+y^2} \ge \cosh(y)\\ $$

Substitute $y$ by $\sinh(x)$ and notice $\sqrt{1+y^2} = \cosh(x)$, this reduces to our desired inequality:

$$\sinh(\cosh(x)) \ge \cosh(\sinh(x))$$

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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 $x=0$.

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  • $\begingroup$ (+1) Very nice. Since $\cosh(x)\gt0$, we don't need to worry about signs when taking square roots. $\endgroup$ – robjohn May 27 '15 at 4:22
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Let $t=\sinh x$. Now we can square the inequality and instead try proving $$\sinh^2(\sqrt{1+t^2})=\sinh^2(\cosh x) \ge \cosh^2(\sinh x)=1+\sinh^2t$$

So it is enough to show $f(t) = \sinh^2(\sqrt{1+t^2})-\sinh^2t-1 \ge 0$. As $f$ is even and $f(0)> 0$, it is enough to show it is increasing for positive $t$. Hence we look at $$f'(t) = \frac{t\sinh (2\sqrt{1+t^2})}{\sqrt{1+t^2}} - \sinh(2t)$$ To show this is positive, it suffices to note by differentiating that the function $g(t) = \dfrac{\sinh t}{t}$ is increasing, so $g(2\sqrt{1+t^2})> g(2t)$. Hence proved...

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For $x\ge0$, the Mean Value Theorem says that for some $\sinh(x)\lt\xi\lt\cosh(x)$, $$ \begin{align} \sinh(\cosh(x))-\sinh(\sinh(x)) &=\cosh(\xi)(\cosh(x)-\sinh(x))\\ &\gt\cosh(\sinh(x))\,e^{-x}\tag{1} \end{align} $$ Furthermore, $$ \cosh(\sinh(x))-\sinh(\sinh(x))=e^{-\sinh(x)}\tag{2} $$ Therefore, subtracting $(2)$ from $(1)$, then applying $\cosh(x)\ge1$ and $\sinh(x)\ge x$, we get $$ \begin{align} \sinh(\cosh(x))-\cosh(\sinh(x)) &\gt e^{-x}\cosh(\sinh(x))-e^{-\sinh(x)}\\ &\ge e^{-x}-e^{-\sinh(x)}\\ &\ge0\tag{3} \end{align} $$ Since $\sinh(\cosh(x))-\cosh(\sinh(x))$ is even, $(3)$ implies that strict inequality holds for all $x$: $$ \sinh(\cosh(x))\gt\cosh(\sinh(x))\tag{4} $$

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