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
 A: 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 $\triangle ABC$.$\qquad\square$
A: Let $(a,b,c) = (\angle A,\angle B,\angle C)$ in $\triangle ABC$
Let $Γ$ be the circumcircle of $\triangle ABC$ with centre $O$
Then $A,O$ are on the same side of $BC$ because $\triangle ABC$ is acute
Let $D,E$ be the intersections of the perpendicular bisector of $BC$ with $Γ$ such that $A,D$ are on the same side of $BC$
Then $(M,X;D,E) = (B,C;D,E) = -1$ by the projection through $C$
Let $F,G$ be the second intersections of $AM,AX$ with $Γ$ respectively
Then $(F,G;D,E) = (M,X;D,E) = -1$ by the projection through $A$
Thus $F,G$ are symmetric about $DE$ and hence $\angle MAE = \angle EAX$
Thus $\frac{|AM|}{|AX|} = \frac{|ME|}{|EX|}$
Also for fixed $B,C,Γ$, $r$ is minimum when $|IO|$ is maximized by Euler's Triangle Formula
Now $\angle BIE = \angle BAI + \angle IBA = \angle IAC + \angle CBI = \angle EBC + \angle CBI = \angle EBI$
Thus $|BE| = |IE|$ and by symmetry $|CE| = |IE|$
Thus $|IO|$ is maximized when $I$ is furthest from $DE$ because $I$ is on the circle through $B,C$ centred at $E$ and $O$ is between $D,E$
Thus $|IO|$ is maximized when WLOG $a = c$
If $a = c$:
  $\frac{r}{R} = \sin(\frac{1}{2}a) \ge \sin(30^\circ) = \frac{1}{2}$
  $|MX| = |MC| \tan(\angle BCX) = |MC| \tan(a)$
  $|MC| = |ME| \tan(\angle CED) = |ME| \cot(\angle BCE) = |ME| \cot(\frac{1}{2}a)$
  $\frac{|EX|}{|ME|} = \frac{|MX|}{|ME|} - 1 = \tan(a) \cot(\frac{1}{2}a) - 1 = \frac{2\sin(\frac{1}{2}a)\cos(\frac{1}{2}a)}{\cos(a)} \cdot \frac{\cos(\frac{1}{2}a)}{\sin(\frac{1}{2}a)} - 1 = \frac{1}{\cos(a)} \ge \frac{1}{\cos(60^\circ)}$
    $ = 2$
  Thus $\frac{r}{R} \ge \frac{|ME|}{|EX|}$
Therefore $\frac{r}{R} \ge \frac{|ME|}{|EX|}$ in all cases
