Proving that $\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$ I'm trying to prove the inequality :
$\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$
I played a bit with Geogebra and it looks true. I've tried proving it via convexity but so far I'm unsuccesful. 
Thank you for your help.
 A: By AM-GM $$\cosh x\cosh y=\left(1+\frac{x^2}{2}+\frac{x^4}{24}+...\right)\left(1+\frac{y^2}{2}+\frac{y^4}{24}+...\right)\geq$$
$$\geq\left(1+\frac{x^2}{2}\right)\left(1+\frac{y^2}{2}\right)\geq1+\frac{x^2+y^2}{2}\geq2\sqrt{1\cdot\frac{x^2+y^2}{2}}\geq\sqrt{x^2+y^2}$$
A: By Taylor's theorem with the Lagrange form of the remainder,
$$ \cosh{x} = 1 + \frac{x^2}{2}\cosh{\xi} $$
for some $\xi$ with $0 \leq \lvert\xi\rvert \leq \lvert x\rvert$. Hence $\cosh{x} \geq 1+\frac{1}{2}x^2$. Thus
$$ \cosh{x}\cosh{y} \geq (1+\tfrac{1}{2}x^2)(1+\tfrac{1}{2}y^2) \geq 1 + \frac{1}{2}(x^2+y^2), $$
since $x^2y^2>0$.
Now by the AM–GM inequality $\frac{1}{2}(a+b) \geq \sqrt{ab}$, taking $a=2$ and $b=x^2+y^2$,
$$1+\tfrac{1}{2}(x^2+y^2) \geq \sqrt{2(x^2+y^2)} > \sqrt{x^2+y^2}, $$
as required.
A: (1) In surface of constant curvature $-1$, cosine law is $$ \cosh\
x\cosh\ y =\cosh\ z$$ where $\frac{\pi}{2}$ is corresponded angle
wrt side whose length is $z$.
So $z\geq \sqrt{x^2+y^2}$ so that we have a claim that $\cosh\ z\geq
z$.
(2) $\cosh\ z-z = 1+ \frac{z^2}{2} + \frac{z^4}{4!} +\cdots -z
 =e^{-z} + \{
 \frac{z^3}{3!} +\frac{z^5}{5!}+\cdots \} >0 $
