A geometric inequality

Question

Let $M$ be a point inside the triangle $ABC$. $AM$ intersects the circumcircle of $MBC$ for the second time at $D$. Analogously define $E,F$. Prove the following : $$ \frac{AD}{MD}+\frac{BE}{ME}+\frac{CF}{MF}\ge \frac{9}{2}$$ I solved this when $M=O$ the circumcenter of $ABC$ using an Inversion $\mathcal{I}(O,R)$ in the circumcircle of $ABC$. Which works out nicely. But it is not working out in the general case.

Answer 1

The inequality $\iff \dfrac{AM}{MD}+\dfrac{BM}{ME}+\dfrac{CM}{MF}\ge \dfrac{3}{2}$

denote $X,Y,Z$ is the circumcenter .connect $XM$ and corss $ZY$ at $G$

it is trivial $AG=GM,MX=XD，\angle GMA=\angle XMD \implies \dfrac{AM}{MD}=\dfrac{GM}{MX}$

so the problem become in $\triangle XYZ$, there is $M ,\sum_{cyc}\dfrac{GM}{MX} \ge \dfrac{3}{2}$

now look at $\triangle XYZ$ in details:

$d_x=MR\perp YZ=x,d_y=MS\perp XZ=y,d_z=MR\perp YX=z,h_x=XH \perp YZ=x$

$\dfrac{GM}{MX}=\dfrac{d_x}{h_x-d_x},\sum_{cyc}\dfrac{GM}{MX}=\sum_{cyc}\dfrac{d_x}{h_x-d_x} \ge \dfrac{3}{2} \iff \sum_{cyc}\dfrac{h_x}{h_x-d_x} \ge \dfrac{9}{2}$

$\dfrac{h_x}{h_x-d_x}=\dfrac{1}{1-\dfrac{d_x}{h_x}}=\dfrac{1}{1-\dfrac{x*d_x}{x*h_x}}=\dfrac{1}{1-\dfrac{S_{MYZ}}{S_{XYZ}}} \implies \sum_{cyc}\dfrac{h_x}{h_x-d_x}=\sum_{cyc}\dfrac{1}{1-\dfrac{S_{MYZ}}{S_{XYZ}}} \ge \dfrac{(1+1+1)^2}{3-\sum_{cyc}\dfrac{S_{MYZ}}{S_{XYZ}}}=\dfrac{9}{2}$

becasue $\sum_{cyc}\dfrac{S_{MYZ}}{S_{XYZ}}=1$

The "=" will hold when $\dfrac{d_x}{h_x}=\dfrac{d_y}{h_y}=\dfrac{d_z}{h_z}=\dfrac{1}{3}$

QED.

BTW, for $\triangle ABC$ there is two $M$ satisfy the "=". but I haven't a solid proof which may be another interesting question.