Maximum area of a triangle defined by a point $P$ inside another triangle (Try using ceva's theorem)

Question

Let $P$ be a point inside an acute triangle $\triangle ABC$ and let $D$, $E$, and $F$ be the points of intersection of the lines $AP$, $BP$ and $CP$ with sides $BC$, $CA$ and $AB$, respectively. Determine $P$ so that the area of ​​the triangle $\triangle DEF$ is maximal.

Answer 1

Let $\frac{AF}{FB}=x$, $\frac{BD}{DC}=y$ and $\frac{CE}{EA}=z$.

Thus, by Cheva's theorem $xyz=1$ and $$\frac{S_{\Delta AFE}}{S_{\Delta ABC}}=\frac{AF\cdot AE}{AB\cdot AC}=\frac{x}{(1+x)(1+z)},$$ $$\frac{S_{\Delta BFD}}{S_{\Delta ABC}}=\frac{BD\cdot BF}{BA\cdot BC}=\frac{y}{(1+y)(1+x)},$$ $$\frac{S_{\Delta CDE}}{S_{\Delta ABC}}=\frac{CD\cdot CE}{CB\cdot CA}=\frac{z}{(1+z)(1+y)}$$ and by AM-GM $$\frac{S_{\Delta FDE}}{S_{\Delta ABC}}=1-\sum_{cyc}\frac{x}{(1+x)(1+z)}=$$ $$=\frac{\prod\limits_{cyc}(1+x)-\sum\limits_{cyc}(x+xy)}{\prod\limits_{cyc}(1+x)}=\frac{2}{\prod\limits_{cyc}(1+x)}\leq\frac{2}{8\sqrt{xyz}}=\frac{1}{4}.$$ The equality occurs for $x=y=z=1$, which gives that $P$ is is an intersect point of medians of the triangle.