I didn't find a synthetic proof but here's a way to do this analytically: Claim 1:(Well Known) $\frac{AM}{AX} = \cos (A)$ Claim 2:(well Known) $\cot A = \frac{b^2+c^2-a^2}{4 \delta} $ also note it suffices to prove $r \geq acot{(A)}$, substitute claim 2 and we are done.

Let $P$ be a $A-Humpty$ point. By inversion+symmetry onto angle bisector of $BAC$ we have $\frac{AM}{AX}=\frac{AM\cdot AP}{AB\cdot AC} = \frac{AM \cdot (AM-MP)}{AB\cdot AC} = \frac{AM \cdot \left( AM-\frac{MC^2}{MA} \right)}{AB\cdot AC} = \frac{AM^2-CM^2}{AB\cdot AC} = \frac{AB^2+AC^2-BC^2}{2AB\cdot AC} = cos A$. So, we need to prove that $r \geq RcosA$. Let $AB=c,BC=a,CA=b$. We have $r=\frac{b+c-a}{2} \cdot \frac{sinA}{1+cosA}, RcosA=\frac{a\cdot cosA}{2sinA}$. So, by law of sines we need to check that $\frac{cosA+1}{sin^2A} \leq \frac{sinB+sinC}{sinA}$. Put $A=180^{\circ}-B-C$, then we need to check that $-cos(B+C)+1 \leq sin(B+C)(sinB+sinC)$ which is equivalent to $2 sin^2((B + C)/2) (cosB + cosC - 1) \geq 0$. And the last is true because $A \leq 60^{\circ}$, so $cosA \leq 1/2$ and it is well-known that $cosA+cosB+cosC \geq \frac{3}{2}$, so we are done!

What is this problem?

starchan wrote: What is this problem?

Actually, $\cos A + \cos B + \cos C \leq 3/2$ so your solution is incorrect. I fix it like that : We need to prove $\cos B+\cos C \geq 1$. Let $AH$ be the altitude of the triangle $ABC$. We have $\cos B = BH/AB \geq BH/BC$ and $\cos C=CH/AC \geq CH/BC$ so $\cos B+\cos C\geq 1$.

what is interesting, is that this problem dates at least from 2005, as mentioned here