Meta-Medians

Let's introduce $a=BC, b=CA, c=AB, \alpha=CF, \beta=AD, \gamma=BE$ . We want to find

$\frac{\alpha^2+\beta^2+\gamma^2}{a^2+b^2+c^2}$

Law of cosines applied three times to $\triangle ABC$ :

$a^2=b^2+c^2-2bc\times cosA$

$b^2=a^2+c^2-2ac\times cosB$

$c^2=b^2+a^2-2ab\times cosC$

The three lines added together and the result simplified:

$a^2+b^2+c^2=2(a^2+b^2+c^2)-(2bc\times cosA+2ac\times cosB+2ab\times cosC)$

$a^2+b^2+c^2=(2bc\times cosA+2ac\times cosB+2ab\times cosC)$

Law of cosines used for the not-quite-medians:

$\alpha^2=b^2+(\frac{c}{3})^2-2b\frac{c}{3}\times cosA$

$\beta^2=c^2+(\frac{a}{3})^2-2c\frac{a}{3}\times cosB$

$\gamma^2=a^2+(\frac{b}{3})^2-2a\frac{b}{3}\times cosC$

There three lines added and simplified:

$\alpha^2+\beta^2+\gamma^2=\frac{10}{9}(a^2+b^2+c^2)-\frac{1}{3}(2bc\times cosA+2ac\times cosB+2ab\times cosC)$

The last expression replaced with the result from the first three lines:

$\alpha^2+\beta^2+\gamma^2=\frac{10}{9}(a^2+b^2+c^2)-\frac{1}{3}(a^2+b^2+c^2)=\frac{7}{9}(a^2+b^2+c^2)$