Area Inequality in Triangle

The following problem and its solution have been posted by Leo Giugiuc at the CutTheKnot facebook page.

$D,E,F$ are second intersections of the medians of $\Delta ABC$ with the circumcircle $(ABC).$

Let $[X]$ denote the area of figure $X.$ Prove that $[\Delta ABC]\le [\Delta DEF].$

Solution

Below, as usual, $a,b,c$ denote the side lengths of $BC,AC,AB,$ respectively;, $m_{a},m_{b},m_{c}$ are the corresponding medians.

Let $M$ be the midpoint of $BC.$ then, by the Intersecting Chords Theorem, $BM\cdot CM=AM\cdot DM,$ so that

$\displaystyle DM=\frac{a^{2}}{4m_{a}}.$

Since $\displaystyle m_{a}=\frac{b^{2}+c^{2}}{2}-\frac{a^{2}}{4},$ we get

$\displaystyle DG=\frac{m_{a}}{3}+\frac{a^{2}}{4m_{a}} = \frac{a^{2}+b^{2}+c^{2}}{6m_{a}}.$

Similarly, $\displaystyle EG== \frac{a^{2}+b^{2}+c^{2}}{6m_{b}}.$ Now, triangles $GBA$ and $GDE$ are similar so that $\displaystyle\frac{BG}{DG}=\frac{c}{DE}.$ A combination of the above gives

$\displaystyle DE=\frac{c(a^{2}+b^{2}+c^{2})}{4m_{a}m_{b}}.$

By analogy, $\displaystyle EF=\frac{a(a^{2}+b^{2}+c^{2})}{4m_{b}m_{c}}$ and $\displaystyle DF=\frac{b(a^{2}+b^{2}+c^{2})}{4m_{a}m_{c}}.$

As we know, for a triangle $ABC$ with sides $a,b,c$ and the circumradius $R,$ $[\Delta ABC]=\displaystyle\frac{abc}{4R}.$ Similarly, for $\Delta DEF,$

$[DEF]=\displaystyle\frac{1}{4R}\frac{abc(a^{2}+b^{2}+c^{2})^{3}}{64m_{a}^{2}m_{b}^{2}m_{c}^{2}}.$

Thus, to prove $[\Delta DEF]\ge [\Delta ABC],$ suffice it to show that $(a^{2}+b^{2}+c^{2})^{3}\ge 64m_{a}^{2}m_{b}^{2}m_{c}^{2}.$

Denote $a^{2}=x, b^{2}+y,c^{2}=z, s=x+y+z.$ We have to prove that $s^{2}\ge(2s-3x)(2s-3y)(2s-3z).$ But this follows from the Arithmetic Mean - Geometric Mean inequality:

\begin{align} \displaystyle (2s-3x)(2s-3y)(2s-3z) &\le \bigg(\frac{(2s-3x)+(2x-3y)+(2s-3z)}{3}\bigg)^{3}\\ &=(x+y+z)^{3}. \end{align}