Two inequalities in a triangle

Question

I'm trying to prove that in a triangle with side lengths $a,b,c$, median lengths $m_a, m_b, m_c$ and circumdiameter $D$ the following inequality holds: $$ \frac{a^2+b^2}{m_c}+\frac{b^2+c^2}{m_a}+\frac{c^2+a^2}{m_b}\le 6D $$ In order to prove it, I tried to use $m_a\ge h_a$ etc. which was when I discovered that the following inequality should be true as well: $$ \frac{a^2+b^2}{h_c}+\frac{b^2+c^2}{h_a}+\frac{c^2+a^2}{h_b}\ge 6D $$ Where $h_a, h_b, h_c$ are the lengths of the hights of the triangle. My question is: how can we prove these inequalities?

Answer 1

For the second one :

$ \frac{a^2+b^2}{h_c} = \frac{a^2+b^2}{\frac{2S}{c} } = \frac{c(a^2+b^2)}{2S} \geq \frac{2abc}{2S}= \frac{abc}{S} =4R$ because $a^2+b^2 \geq 2ab$ and $4R=\frac{abc}{S} $

analogously you prove for other 2 terms so we get :

$\sum_{cyc} \frac{a^2+b^2}{h_c} \geq 3 \cdot 4R = 12R = 6D $

For the other one:

Let $AD$ be median and it intersects circumcenter at $D'$.

Applying power of a point on point $D$ we get $AD \cdot DD' = BD \cdot DC \Rightarrow DD'= \frac{a^2}{4m_a}$ because $BD=DC=\frac{a}{2}$

Since $AD' \leq 2R \Rightarrow AD+DD' \leq 2R$ we get

$m_a+\frac{a^2}{4m_a} \leq 2R \Rightarrow \frac{4m_a^2+a^2}{4m_a} \leq 2R$ and putting $m_a^2=\frac{2b^2+2c^2-a^2}{4}$ we get $\frac{b^2+c^2}{m_a}\leq 4R $

And finally $\sum_{cyc} \frac{b^2+c^2}{m_a} \leq 3 \cdot 4R =12R = 6D $