A, B and C are the vertices of a triangle. Denote $m_a$, $m_b$ and $m_c$ the medians from A, B and C. Prove the inequality: $$\sum_{cyc}{a^2bcm_a}\geq\sum_{cyc}{cS(a^2+b^2)}$$where a, b and c are the side lengths BC, CA and AB respectively and S is the area of the triangle. This can be rewritten as $$\sum_{cyc}{am_a}\geq S(\sum_{cyc}{\frac{a^2+b^2}{ab}})$$ Any ideas? Thanks!

  • 3
    $\begingroup$ Where did you get this? $\endgroup$ – Sawarnik Mar 7 '14 at 22:10

Since $S=\frac{abc}{4R}$ and $2m_a^2=b^2+c^2-\frac{a^2}{2}$, we can put the given inequality in the following form: $$4R\sum_{cyc}a m_a \geq \sum_{cyc} a(b^2+c^2) = \sum_{cyc}a(2m_a^2+a^2/2),$$ $$\sum_{cyc}a(2Rm_a - m_a^2 - a^2/4) \geq 0,$$ $$\sum_{cyc} a\cdot(m_a-R)^2 \leq \sum_{cyc}a\cdot\left(R^2-\frac{a^2}{4}\right).\tag{1}$$ Let $O$ be the circumcenter of $ABC$ and $\Gamma_A$ the circle having center $A$ and radius $R$: obviously $O\in\Gamma_A$. Let $M_A$ be the midpoint of $BC$: $OM_A$ is orthogonal to $BC$ and $OM_A^2 = R^2-\frac{a^2}{4}$ by Pythagoras' theorem. Now $$ |R-m_a|=d(M_A,\Gamma_A)\leq d(M_A,O) = \sqrt{R^2-\frac{a^2}{4}},$$ and $(1)$ follows.

  • $\begingroup$ Would you explain how $(1)$ is equivalent to the inequality we want? $\endgroup$ – robjohn Mar 13 '14 at 21:18
  • $\begingroup$ @robjohn: I added two explicatory lines. $\endgroup$ – Jack D'Aurizio Mar 13 '14 at 21:56
  • $\begingroup$ $4R\sum\limits_\text{cyc} am_a\ge\sum\limits_\text{cyc}a(b^2+c^2)$ is the inequality that I am not sure how we are to see as equivalent to the given inequality. $\endgroup$ – robjohn Mar 13 '14 at 22:34
  • $\begingroup$ @robjohn: Yes it is, since $abc = 4RS$. $\endgroup$ – Jack D'Aurizio Mar 13 '14 at 23:00
  • $\begingroup$ And $\sum_{cyc}c(a^2+b^2) = \sum_{cyc}a(b^2+c^2)$, obviously. $\endgroup$ – Jack D'Aurizio Mar 13 '14 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy