part a) let the intersection of AB and CM be F. There is a formula for the length of the median which can be derived by constructing a parallelogram; we use this to find $FC^2=\frac{2a^2+2b^2-c^2}{4}$ We can also use power of a point with regard to point X; $FA^2=\frac{c^2}{4}=FM\cdot FC=\frac{FC^2}{3}=\frac{2a^2+2b^2-c^2}{12}$ Thus, $2a^2+2b^2=4c^2$ or $a^2+b^2=2c^2$. This looks a lot like the law of cosines; $c^2=a^2+b^2-2ab\cos C\implies c^2=2ab\cos C$ Also, we can now use the median formula to find the length of the other two medians. The median from A, $m_a^2=\frac{3}{4}b^2$ and $m_b^2=\frac{3}{4}a^2$ Now we can directly solve for what we want. $\frac{1}{2}m_ab\sin CAM=\frac{[ABC]}{2}=\frac{1}{2}m_ba\sin CBM$ $\sin CAM+\sin CBM=[ABC](\frac{1}{m_b\cdot a}+\frac{1}{m_a\cdot b})=\frac{1}{\sqrt{3}}\cdot ab\sin C\cdot\frac{a^2+b^2}{a^2b^2}$ Now, we can substitute $ab$ for $\frac{c^2}{2\cos C}$ to get $\frac{1}{\sqrt{3}}(a^2+b^2)\sin C\cdot\frac{2\cos C}{c^2}=\frac{2}{\sqrt{3}}\sin 2C\ge \frac{2}{\sqrt{3}}$

What is the (b) part? Anyone?