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Not long ago, I asked how to prove that $\cosh(x)\cosh(y) \geq \sqrt{x^2 + y^2}$. People told me to use Taylor expansions and the Arithmetic-Geometric Inequality.

Now I'm trying to prove that $\cosh^{-1} (\cosh(x)\cosh(y) ) \geq \sqrt{x^2 + y^2}$ with similar techniques but I'm once again unsuccessful.

Thank you for your help.

In case anybody is interested, I'm actually trying to prove that the upper-half plane verifies the CAT(0) inequality by hand. This means I'm trying to do it without the classical result that if $k<k'$, CAT(k) implies CAT(k'). In order to do this, I thought that treating the special case of a right triangle could be useful, and the above inequality will give me this result.

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First note that $$\cosh(x)\cosh(y)=\frac{1}{2}\big(\cosh\left(x+y\right)+\cosh\left(x-y\right)\big)\,.$$ Observe that the function $f(t):=\cosh\big(\sqrt{t}\big)$, defined for $t\geq 0$, is convex (this is due to the fact that $\tanh(a)\leq a$ for all $a\geq 0$). Hence, using Jensen's Inequality, $$\cosh(x)\cosh(y)=\frac{1}{2}\Big(f\left((x+y)^2\right)+f\left((x-y)^2\right)\Big)\geq f\left(\frac{(x+y)^2+(x-y)^2}{2}\right)\,,$$ so we have $$\cosh(x)\cosh(y)\geq \cosh\left(\sqrt{x^2+y^2}\right)\,.$$

Note that $f''(t)=f(t)\,\left(\frac{\sqrt{t}-\tanh\left(\sqrt{t}\right)}{4t\sqrt{t}}\right)$.

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