Let $G$ be the point of intersection of the medians in the triangle $ABC$. Let us denote $A_1, B_1, C_1$ the second points of intersection of lines $AG, BG, CG$ with the circle circumscribed around the triangle. Prove that $AG + BG + CG \le A_1C + B_1C + C_1C$.
(Yasinsky V.A.)
Let $a, b, c$ be the respective sidelengths and $m_a, m_b, m_c$ be the lengths of the respective medians. Then $AG=\frac{2m_a}{3}$ etc. and $A_1G=\frac{a^2}{4m_a}+\frac{m_a}{3}$ etc. and we need to prove that
$$\sum_{cyc}\frac{m_a}{3}\le \sum_{cyc}\frac{a^2}{4m_a}\iff \frac{1}{12}\sum_{cyc}\frac{2b^2+2c^2-4a^2}{m_a}\le 0\iff$$$$\sum_{cyc}(c^2-a^2)\left(\frac{1}{m_a}-\frac{1}{m_c}\right)\le 0\iff \sum_{cyc}\frac{c^2-a^2}{m_am_c}\cdot \frac{a^2-c^2}{m_c+m_a}\le 0.$$