<s>[INST] 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.
 [/INST] 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.
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