<s>[INST] Areas in triangle 
In the above isosceles right triangle ABC, with its two sides $AB = AC = 1$ unit,  we take a random point D on the hypotenuse and draw perpendicular lines to the sides AB and AC, which intersect them at points E and F respectively.
Show that the maximum of the three areas AFDE, EBD and CDF is always $\geq \frac {2}{9}$.
If we set $EB = x$ then, area $EBD = \frac {x^2}{2}$,
$AFDE = (1-x)*x$,
$CDF = \frac {(1-x)^2}{2}$.
Also $AFDE + EBD + CDF = \frac {1}{2}$.
Clearly the 3 areas can't be equal.
We can only have 2 of them equal, when $x=0.5$ or $x=0.33$ or $x=0.66$.
When $x=0.33 = \frac {1}{3}$ then
$EBD = \frac {x^2}{2} = \frac{1}{18}$ and $AFDE = CDF = \frac{1}{2}*(\frac{9}{18}-\frac{1}{18}) = \frac{2}{9}$.
But I don't know if this is a sufficient proof.
 [/INST] Start with D as the mid point of BC. WLOG, imagine D moves up by a vertical distance of $\Delta x$ .
\begin{align}S_{AEDF}=\frac{1}{4}-\Delta x^2, S_{\triangle DFC}=\frac{1}{2}(\frac{1}{2}+\Delta x)^2\end{align}
\begin{align}S_{AEDF}\geq \frac{2}{9}(\Delta x \leq\frac{1}{6})\\
 S_{\triangle DFC}\geq \frac{2}{9}(\Delta x\geq \frac{1}{6})\end{align}
Similar conclusion can be obtained if D moves down by $\Delta x$.

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