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.

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.
