Prove that $\sinh(\cosh(x)) \geq \cosh(\sinh(x))$ 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?
 A: 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...
A: 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 $x$:
$$
\sinh(\cosh(x))\gt\cosh(\sinh(x))\tag{4}
$$
A: 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 $x=0$.
A: 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:
$$\sinh(\cosh(x)) \ge \cosh(\sinh(x))$$
A: Substitute :
$e^y=x$
We get :
$$\sinh\left(\frac{\left(x+\frac{1}{x}\right)}{2}\right)\geq \cosh\left(\left(x-\frac{1}{x}\right)\cdot0.5\right)$$
Now substitute again we get :
$$\left(x+\frac{1}{x}\right)0.5-\frac{1}{\left(\left(x+\frac{1}{x}\right)\cdot0.5\right)}-\left(\left(x-\frac{1}{x}\right)0.5+\frac{1}{\left(x-\frac{1}{x}\right)0.5}\right)\geq 0$$
Or :
$$\frac{\left(3x^{4}+1\right)}{\left(1-x\right)x\left(x+1\right)\left(x^{2}+1\right)}\geq0$$
The conclusion is smooth .
A: We can actually prove strict inequality.
Let's start with the observation that, since $\cosh u\ge1\gt0$ for all $u$ and $\sinh u\gt u\gt0$ for all $u\gt0$, we have
$$\begin{align}
\sinh(\cosh x)\gt\cosh(\sinh x)
&\iff\sinh^2(\cosh x)\gt\cosh^2(\sinh x)\\
&\iff\cosh^2(\cosh x)-1\gt\cosh^2(\sinh x)\\
&\iff\cosh^2(\cosh x)-\cosh^2(\sinh x)\gt1\\
&\iff(\cosh(\cosh x)+\cosh(\sinh x))(\cosh(\cosh x)-\cosh(\sinh x))\gt1
\end{align}$$
Now $\cosh(\cosh x)+\cosh(\sinh x)\ge1+1=2$, so it suffices to prove that
$$\cosh(\cosh x)-\cosh(\sinh x)\gt{1\over2}$$
But $\cosh(a+b)-\cosh(a-b)=2\sinh a\sinh b\gt2ab$ (with the inequality requiring $a,b\gt0$), we have, letting $a=e^x/2$ and $b=e^{-x}/2$ (both of which are clearly positive),
$$\cosh(\cosh x)-\cosh(\sinh x)\gt2\cdot{e^x\over2}\cdot{e^{-x}\over2}={1\over2}$$
as desired.
