Given concurrent cevians $AD$, $BE$, $CF$ of $\triangle ABC$, show $\text{area }\triangle DEF\leq \tfrac14\text{}\triangle ABC$ 
Let $M$ be a point in $\triangle ABC$. Lines $AM$, $BM$, $CM$ respectively intersect $BC$, $CA$, $AB$ at $D$, $E$, $F$. Prove that 
  $$S_{DEF}\leq \tfrac{1}{4}S_{ABC}$$
  where $S_{XYZ}$ indicates the area of $\triangle XYZ$.


Let the segment ratios be 
$$x= \frac {CD}{DB}, \>\>\> y= \frac {AE}{EC}, \>\>\> z= \frac {BF}{FA}$$
According to the Ceva's theorem,
$$xyx=1\tag{1}$$
Evaluate the area ratio
$$\frac{S_{AEF}}{S_{ABC}} = \frac{ \frac 12 AE\cdot AF\sin A}{\frac 12 AB\cdot AC\sin A}
=\frac{1}{(1+z)(1+\frac 1y)}=\frac{y}{(1+z)(1+y)}$$
Similarly,
$$\frac{S_{BDF}}{S_{ABC}} = \frac{z}{(1+x)(1+z)}, \>\>\>\>\> \frac{S_{CDE}}{S_{ABC}} = \frac{x}{(1+y)(1+x)}$$
Then, 
$$\frac{S_{DEF}}{S_{ABC}} = 1 - \frac{S_{AEF} + S_{BDF}+S_{CDE} }{S_{ABC}} $$
$$= 1- \frac{y}{(1+z)(1+y)}-\frac{z}{(1+x)(1+z)}-\frac{x}{(1+y)(1+x)}$$
Simplify the expression,
$$\frac{S_{DEF}}{S_{ABC}} = \frac{2}{(1+x)(1+y)(1+z)} $$
Apply the AM-GM inequality to the denominator,
$$(1+x)(1+y)(1+z)=1+x+y+z+xy+yz+zx+xyz \ge 8[(xyz)^4]^{\frac 18}=8$$ 
Thus 
$$\frac{S_{DEF}}{S_{ABC}} \le \frac 14$$
