Problem 1137

Proof As shown in Figure 6, let the center of circle $\Gamma$ be $O$, and $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$ respectively. The circle $\Gamma$ is tangent to the three semicircles at points $D'$, $E'$, $F'$, and the radii of these three semicircles are $d'$, $e'$, $f'$ respectively. Then $DD'$, $EE'$, $FF'$ all pass through point $O$, and $p=d'+e'+f'$.
$$\begin{array}{l} \text{Let } d=\frac{p}{2}-d'=\frac{-d'+e'+f'}{2}, \\ e=\frac{p}{2}-e'=\frac{d'-e'+f'}{2}, \\ f=\frac{p}{2}-f'=\frac{d'+e'-f'}{2}. \end{array}$$

Then $d+e+f=\frac{p}{2}$.
Inside $\triangle ABC$, semicircles are drawn with $D$, $E$, $F$ as centers and $d$, $e$, $f$ as radii respectively.
Since $d+e=f'=\frac{1}{2} AB=DE$,
$$\begin{array}{l} e+f=d'=\frac{1}{2} BC=EF, \\ f+d=e'=\frac{1}{2} AC=FD, \end{array}$$

Therefore, these three smaller semicircles are pairwise tangent, and these points of tangency are the points where the incircle of $\triangle DEF$ is tangent to its three sides.

Let $DD'$, $EE'$, $FF'$ intersect the smaller semicircles at points $D''$, $E''$, $F''$ respectively.

Since these semicircles do not overlap, and $O$ is a point outside these semicircles, we have $D'O > D'D''$, i.e., $t > \frac{p}{2}$.
Thus, the left side of equation (9) is proved.
Let $g=t-\frac{p}{2}$. Then
$$OD''=OE''=OF''=g.$$

Therefore, the circle with center $O$ and radius $g$ is tangent to these three mutually tangent semicircles.
By Descartes' theorem,
$$\begin{array}{l} 2\left(\frac{1}{d^2}+\frac{1}{e^2}+\frac{1}{f^2}+\frac{1}{g^2}\right)=\left(\frac{1}{d}+\frac{1}{e}+\frac{1}{f}+\frac{1}{g}\right)^2 \\ \Rightarrow \frac{1}{g}=\frac{1}{d}+\frac{1}{e}+\frac{1}{f}+2 \sqrt{\frac{d+e+f}{def}}. \\ \text{Also, } S_{\triangle DEF}=\frac{1}{4} S_{\triangle ABC}=\frac{pr}{4} \\ =\sqrt{(d+e+f)def}, \\ \text{Then } \frac{r}{2}=\frac{2}{p} \sqrt{(d+e+f)def}=\sqrt{\frac{def}{d+e+f}}. \end{array}$$

Hence, to prove the right side of equation (9), it is equivalent to proving
$$\frac{r}{2g} \geqslant \frac{1}{2-\sqrt{3}}=2+\sqrt{3}.$$

Notice that,
$$\begin{array}{l} \frac{r}{2g}=\sqrt{\frac{def}{d+e+f}}\left(\frac{1}{d}+\frac{1}{e}+\frac{1}{f}+2 \sqrt{\frac{d+e+f}{def}}\right) \\ =\frac{x+y+z}{\sqrt{xy+yz+zx}}+2, \end{array}$$

where $xd=1$, $ye=1$, $zf=1$.
Thus, it suffices to prove
$$\begin{array}{l} (x+y+z)^2 \geqslant 3(xy+yz+zx). \\ \text{Also, } (x+y+z)^2-3(xy+yz+zx) \\ =\frac{1}{2}\left[(x-y)^2+(y-z)^2+(z-x)^2\right] \\ \geqslant 0, \end{array}$$

Therefore, equation (9) is established.
Thus, the right side of equation (9) is also proved.