IMO shortlist problem In triangle $ABC$, $M$ is an interior point. Segments $AD$, $BE$ and $CF$ are drawn passing through $M$ such that $D$ lies on $BC$, $E$ lies on $AC$,  and $F$ lies on $AB$.  Prove that 
(area of triangle $DEF$)$\le\frac{1}{4} $(area of triangle $ABC$)
I have tried using Ceva's and Menelaus theorem but ended up in so many equations. The coordinate geometry also didn't yield much success. I also would like to know if there can be any use of projective geometry 
 A: OK, it's really messy but here's what you can do. Denote:
$$AF=x$$ $$BD=z$$ $$CE=y $$
Intuition: We want to get rid of one of $x,y,z$ by using Ceva's Theorem and due to symmetry, we want them to alternate (this will ease the calculations)
What you want to show is that 
$$\frac{(b-y)x}{bc}+\frac{z(c-x)}{ac}+\frac{y(a-z)}{ab} \geq \frac{3}{4}$$
by summing the areas of the other $3$ triangles.
If you denote the following:
$$\frac{x}{c}=m$$$$ \frac{y}{b}=n$$$$\frac{z}{a}=p $$
You can the formulate Ceva's Theorem and the claim you want to prove as:
$$mnp=(1-m)(1-n)(1-p)$$
$$Claim: m+n+p-mn-np-mp\geq \frac{3}{4}$$
by just substituting.
Hence, we'd like to show that 
$$mnp\leq \frac{1}{8}$$
holds (using the claim and Ceva's Thm we get this)
Now you can get $p$ in terms of $m,n$ from Ceva's Theorem and plug it in
$$mnp\leq \frac{1}{8}$$
to get 
a quadratic in say $m$.
This quadratic will attain its maximum at $\frac{1}{2}$ and you will be done
(if that problem is true and your calculations are OK, that is :D, didn't finish the calculations)
