<s>[INST] Prove that $m_a\geq \dfrac{b^2+c^2}{4R}$ Let triangle ABC, $m_a$ the lenght of the median from A, $b,c$ the lenghts of the segments AC and AB respectively and R the circumradius. Prove that:
$m_a\geq \dfrac{b^2+c^2}{4R}$.
I found this in a book and the hint was to denote M the midpoint of BC and $A_2$ the second intersection of the median and the circumcircle of the triangle. Then by power of a point we have that $AM \cdot MA_2=a^2/4$. Also $AM+MA_2 \leq 2R$. And then they said this to imply the conclusion.
Please help me understand. Thank you in advance.
 [/INST] As your hint says, from the power of $M$ to $(O)$ we get 
$$AM \cdot MA_2 = BM \cdot MC = \frac{1}{4}a^2$$
$$\Longrightarrow  AA_2 = AM + MA_2 = \frac {AM^2 + \frac {1}{4}a^2}{AM} = \frac {\frac {1}{2}(b^2 + c^2) - \frac {1}{4}a^2 + \frac {1}{4}a^2}{m_a} = \frac {b^2 + c^2}{2m_a}$$
Since a diameter  is greater or equal than the chord $ AA_2,$ we have 
$$2R \ge \frac {b^2 + c^2}{2m_a}$$
and the conclusion follows.
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