<s>[INST] How prove this $\frac{r}{R}\ge\frac{AM}{AX}$ Let acute $\triangle ABC$ have largest angle $\angle A$, and let $R,r$ be the circumradius and inradius respectively. Let $M$ be the midpoint of $BC$, and $X$ be the intersection of the tangents at $B,C$ to the circumcircle of $\triangle ABC$
Show that:
$$\frac{r}{R}\ge\frac{|AM|}{|AX|}$$
Here is the original chinese version:

My method:

So now I have the following problem:
let $x,y,z>0$,and let
$$R=\dfrac{\sqrt{(x^2+y^2+z^2+2xz)(x^2+y^2+z^2-2xz)}}{2y},a=\sqrt{x^2+y^2}$$
$$r=\dfrac{2yz}{2z+\sqrt{x^2+y^2+z^2-2xz}+\sqrt{x^2+y^2+z^2+2xz}}$$
$$b=\sqrt{\dfrac{(x^2+y^2)(x^2+y^2+z^2+2xz)(x^2+y^2+z^2-2xz)}{(x^2+y^2-z^2)^2}}$$
Show that
$$\dfrac{r}{R}\ge\dfrac{a}{b}$$
I think this problem has other methods. If you find other methods, can you explain? 
Thank you very much
 [/INST] Let us identify the plane with the set of complex numbers $\Bbb{C}$, with $\Gamma$ being the unit circle. (So that $R=1$). The complex number representing a point will be denoted by the same, but small, corresponding letter.
Now, $X$ is the image of $M$ under the inversion with respect to $\Gamma$, hence
$$
m=\frac{b+c}{2},\qquad x=\frac{1}{\overline{m}}.
$$ 
Since $|a|=|b|=|c|=1$, we conclude that
$$
AX=|a-x|=|\overline{a}-\overline{x}|=\left\vert\frac{1}{a}-\frac{1}{m}\right\vert=\frac{|a-m|}{|m|}=\frac{AM}{OM}
$$
where $O$ is the center of $\Gamma$. Thus
$$
\frac{AM}{AX}=OM=\cos A \tag{1}
$$
On the other hand, it is well-known that
$$
\frac{r}{R}=4\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{C}{2}  \tag{2}
$$
Thus
$$\eqalign{
\frac{r}{R}-\frac{AM}{AX}&=4\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{C}{2}  -1+2\sin^2 \frac{A}{2}\cr
&=2\sin \frac{A}{2}\left(2\sin \frac{B}{2} \sin \frac{C}{2}+\sin \frac{A}{2}\right)  -1\cr
&=2\sin \frac{A}{2}\left(2\sin \frac{B}{2} \sin \frac{C}{2}+\cos \frac{B+C}{2}\right)  -1\cr
&=2\cos \frac{B+C}{2} \cos \frac{B-C}{2}   -1\cr
&=\cos B+\cos C   -1~\buildrel{\rm(1)}\over{\geq}~ 2\cos\left(\frac{B+C}{2}\right)-1\cr
&\geq2\sin\frac{A}{2}-1~\buildrel{\rm(2)}\over{\geq}~ 2\sin 30^\circ-1=0
}
$$
where $(1)$ follows from the concavity of $x\mapsto \cos x$ on $[0,\pi/2]$, and $(2)$ follows from the fact that $A$ is the largest angle in the acute triangle $\triangle ABC$.$\qquad\square$
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