ratio of lengths of median and altitude

Question

'In a triangle ABC the side opposite $A$ has length $a$, and similarly for $b$ and $c$. Let $l$ be the length of the median and $d$ be the length of the altitude from $A$ to $BC$. Prove that $$\frac{l}{d} \geq \frac{b^2 + c^2}{2bc}.$$

I wrote d^2 in terms of a,b,c using heron's formula = ad/2, and l^2 using Apollonius theorem, but then I get a mess that doesn't seem to simplify. Please help!

Thanks.

Answer 1

Since $l=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$, $d=\frac{2S}{a}$

and $S=\sqrt{p(p-a)(p-b)(p-c)}=\frac{1}{4}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}$, we need to prove that $$abc\sqrt{2b^2+2c^2-a^2}\geq2(b^2+c^2)S$$ or $$4a^2b^2c^2(2b^2+2c^2-a^2)\geq(b^2+c^2)^2\sum\limits_{cyc}(2a^2b^2-a^4)$$ or $$(b^2-c^2)^2(b^2+c^2-a^2)^2\geq0.$$ Done!

Answer 2

$$\dfrac {m_a}{h_a}=\dfrac {2m_aa}{S}=\dfrac {8m_aR}{bc} $$

$$ m_a(2R-m_a) \ge \dfrac {a^2}{4} \Rightarrow 8m_aR \ge \dfrac {b^2+c^2}{2}$$