Prove that $m_a\geq \dfrac{b^2+c^2}{4R}$

Question

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.

Answer 1

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.

Answer 2

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 $$(b^2-c^2)^2(b^2+c^2-a^2)^2\geq0.$$