<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] In the standard notation we need to prove that
$$\frac{1}{2}\sqrt{2b^2+2c^2-a^2}\geq\frac{b^2+c^2}{\frac{abc}{S}}$$ or
$$2abc\sqrt{2b^2+2c^2-a^2}\geq(b^2+c^2)\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ or
$$(b^2-c^2)^2(b^2+c^2-a^2)^2\geq0.$$
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