Geometric Inequality Related To Median, Altitude For a triangle $ABC$, let $m_{a}$, $h_{a}$ be $A$-median, $A$-altitude.
Define $m_{b}$,$h_{b}$ and $m_{c}$,$h_{c}$ likewise.   
Prove that $\dfrac{h_{a}}{m_{b}}+\dfrac{h_{b}}{m_{c}}+\dfrac{h_{c}}{m_{a}}\leq 3$    
I have no solution.
 A: Let $a^2=x$, $b^2=y$ and $c^2=z$.
Thus, by C-S $$\sum_{cyc}\frac{h_a}{m_b}\leq\sqrt{\sum_{cyc}h_a^2\sum_{cyc}\frac{1}{m_b^2}}=\sqrt{\sum_{cyc}\frac{4S^2}{a^2}\sum_{cyc}\frac{4}{2a^2+2c^2-b^2}}=$$
$$=\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)\sum\limits_{cyc}\frac{1}{a^2}\sum_{cyc}\frac{1}{2a^2+2c^2-b^2}}=$$
$$=\sqrt{\frac{\sum\limits_{cyc}(2xy-x^2)\sum\limits_{cyc}xy\sum\limits_{cyc}(2x+2y-z)(2x+2z-y)}{xyz\prod\limits_{cyc}(2x+2y-z)}}=3\sqrt{\frac{\sum\limits_{cyc}(2xy-x^2)\left(\sum\limits_{cyc}xy\right)^2}{xyz\prod\limits_{cyc}(2x+2y-z)}}$$ and it's enough to prove that
$$xyz\prod\limits_{cyc}(2x+2y-z)\geq\sum\limits_{cyc}(2xy-x^2)\left(\sum\limits_{cyc}xy\right)^2$$ or
$$\sum_{sym}(x^4y^2-x^4yz-x^3y^3+x^3y^2z-x^2y^2z^2)\geq0$$ or
$$(x-y)^2(x-z)^2(y-z)^2\geq0$$ and we are done!
A: Denote the side of the triangle opposite vertex $ A $ as $ a $.  Now consider the points on side $ a $ where $ m_a,h_a $ intersect side $ a $.  Label the intersection of $ a $ and $ h_a $ as the point $ X $, and label the intersection of $ a $ and $ m_a $ as the point $ Y $.  Then the triangle $ XYA $ is a right triangle.  The ratio $ \dfrac{h_a}{m_a} $ is actually the sine of the angle that $ m_a $ makes with side $ a $.  Since the sine is always less than or equal to 1, we see that $ \dfrac{h_a}{m_a} \leq 1 $.  The same argument shows that $ \dfrac{h_b}{m_b} \leq 1 $ and $ \dfrac{h_c}{m_c} \leq 1 $.  Taking the sum of all three inequalities, you have your result.
