Let M, K, and L be points on line (AB), (BC), and (CA), respectively. Find the maximum area of smallest the three triangles MAL, KBM, and LCK? Let M, K, and L be points on line (AB), (BC), and (CA), respectively. Find the maximum area of smallest the three triangles  MAL,  KBM, and  LCK in respect to ABC?
I try to guess the answer and then prove it, however it doesn't work to randomly guess.
I draw some pictures, but it hardly tells me anything

We shall denote the area of a triangle $ABC$ with $S_{ABC}$. We shall prove that $\min(S_{MAL}, S_{KBM}, S_{LCK}) \leq \frac{1}{4}S_{ABC}$.
Without loss of generality assume that $S_{MAL}=\max(S_{MAL}, S_{KBM}, S_{LCK})$. Clearly if $\frac{S_{MAL}}{S_{ABC}} \leq \frac{1}{4}$, we are done. Otherwise consider $\frac{S_{MAL}}{S_{ABC}}=\frac{AM}{AB} \cdot \frac{AL}{AC}>\frac{1}{4}$. Let $\frac{AM}{AB}=x, \frac{AL}{AC}=y, 0 \leq x, y \leq 1$. Then $xy>\frac{1}{4}$. Also, we have $\frac{MB}{AB}=1-x, \frac{LC}{AC}=1-y$. Clearly if any of $K, L, M$ are equal to $A, B$, or $C$, then $\min(S_{MAL}, S_{KBM}, S_{LCK})=0 \leq \frac{1}{4}S_{ABC}$. Thus $0<x, y<1$.
Let $K'$ be on $BC$ s.t. $\frac{BK'}{CK'}=\frac{1-y}{1-x}$. Then $$\frac{S_{K'BM}}{S_{ABC}}=\frac{BK'}{BC} \cdot \frac{BM}{AB}=\frac{(1-x)BK'}{BC}=\frac{(1-y)CK'}{BC}=\frac{CK'}{BC} \cdot \frac{LC}{AC}=\frac{S_{LCK'}}{S_{ABC}}$$
If $BK \leq BK'$, then $$\frac{S_{KBM}}{S_{ABC}}=\frac{BK}{BC} \cdot \frac{BM}{AB} \leq \frac{BK'}{BC} \cdot \frac{BM}{AB}=\frac{S_{K'BM}}{S_{ABC}}$$ 
Otherwise $BK>BK'$ so $CK<CK'$, so $$\frac{S_{LCK}}{S_{ABC}}=\frac{CK}{BC} \cdot \frac{LC}{AC}<\frac{CK'}{BC} \cdot \frac{LC}{AC}=\frac{S_{LCK'}}{S_{ABC}}=\frac{S_{K'BM}}{S_{ABC}}$$
In any case, we have $$\min(\frac{S_{KBM}}{S_{ABC}}, \frac{S_{LCK'}}{S_{ABC}}) \leq \frac{S_{K'BM}}{S_{ABC}}=\frac{(1-x)BK'}{BC}=\frac{(1-x)(1-y)}{1-x+1-y}$$
It thus suffices to prove that $\frac{(1-x)(1-y)}{1-x+1-y} \leq \frac{1}{4}$, or equivalently, $\frac{1}{1-x}+\frac{1}{1-y} \geq 4$.
Note that $x>xy>\frac{1}{4}$ so $4x-1>0$. Thus 
\begin{align}
\frac{1}{1-x}+\frac{1}{1-y}-4 & =\frac{1}{1-x}+\frac{4x}{4x-4xy}-4 \\
& >\frac{1}{1-x}+\frac{4x}{4x-1}-4 \\
& =\frac{(4x-1)+4x(1-x)-4(1-x)(4x-1)}{(1-x)(4x-1)} \\
& =\frac{12x^2-12x+3}{(1-x)(4x-1)} \\
& =\frac{3(2x-1)^2}{(1-x)(4x-1)} \\
& \geq 0
\end{align}
Therefore $\min(S_{MAL}, S_{KBM}, S_{LCK}) \leq \frac{1}{4}S_{ABC}$. Furthermore, equality holds when $M, K, L$ are the midpoints of $AB, BC, AC$ respectively, so $\max(\min(S_{MAL}, S_{KBM}, S_{LCK}))=\frac{1}{4}S_{ABC}$.
