Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Please don't understand me in the way that you think I have so much time now to spend it on trivial problems like this, but I fear that the below is the only thing I can contribute right now to Orl's nice ISL/ILL project. So here is my solution of this problem. If ABC is our triangle, and X, Y, Z are the midpoints of its sides BC, CA, AB, then we have the vectorial identities $\overrightarrow{AX}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}$, $\overrightarrow{BY}=\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}$, $\overrightarrow{CZ}=\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}$, so that $\overrightarrow{AX}+\overrightarrow{BY}+\overrightarrow{CZ}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{2}+\frac{\overrightarrow{BC}+\overrightarrow{BA}}{2}+\frac{\overrightarrow{CA}+\overrightarrow{CB}}{2}$ $=\frac{\overrightarrow{BC}+\overrightarrow{CB}}{2}+\frac{\overrightarrow{CA}+\overrightarrow{AC}}{2}+\frac{\overrightarrow{AB}+\overrightarrow{BA}}{2}=\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}$. Thus, the vectors $\overrightarrow{AX}$, $\overrightarrow{BY}$, $\overrightarrow{CZ}$ can be positioned in such a way that they form a triangle. Let DEF be this triangle, where $\overrightarrow{EF}=\overrightarrow{AX}$, $\overrightarrow{FD}=\overrightarrow{BY}$, $\overrightarrow{DE}=\overrightarrow{CZ}$. Then, we have EF || AX, FD || BY, DE || CZ, and $EF=AX=m_a$, $FD=BY=m_b$, $DE=CZ=m_c$. In other words, the sides of the triangle DEF are parallel and equal in length to the medians of triangle ABC. Since EF || AX and FD || BY, the orthogonality of the medians $m_a=AX$ and $m_b=BY$ implies the orthogonality of the sidelines EF and FD of triangle DEF. Thus, this triangle DEF is right-angled at F. This proves part a) of the problem. Since the triangle DEF is right-angled at F, the Pythagoras theorem yields $DE^2=EF^2+FD^2$. But since $EF=m_a$, $FD=m_b$, $DE=m_c$, this rewrites as $m_c^2=m_a^2+m_b^2$, and hence $4m_c^2=4m_a^2+4m_b^2$. Using the well-known formulas $4m_a^2=2b^2+2c^2-a^2$, $4m_b^2=2c^2+2a^2-b^2$, $4m_c^2=2a^2+2b^2-c^2$, this takes the form $2a^2+2b^2-c^2=\left(2b^2+2c^2-a^2\right)+\left(2c^2+2a^2-b^2\right)$, what simplifies to $5c^2=a^2+b^2$. Hence, $5\left(a^2+b^2-c^2\right) = 5\left(a^2+b^2\right)-5c^2 = 5\left(a^2+b^2\right)-\left(a^2+b^2\right) = 4\left(a^2+b^2\right) \geq 8ab$ from the well-known inequality $a^2+b^2 \geq 2ab$. And hence part b) is also proven. Note that equality holds if and only if $a=b=\sqrt{\frac52c}$. Darij