When triangle is divided into four, at least one is not bigger than a quarter of the original proof The problem:
On a $\triangle ABC$ the points $M, K, L$ are chosen respectively on the sides $AB, BC, CA$. Prove that the area of at least one of $\triangle AML, \triangle BMK, \triangle CKL$ will be less than or equal to a quarter of the area of $\triangle ABC$.
I know that the biggest possible they all can be without any being less than ${1\over4}$ is if they are all the same and equal to one quarter. It seems intuitive, that if any of the points is moved, the size of at least one of them will decrease, but I have troubles trying to prove it mathematically.
Any help would be greatly appreciated. Thank you in advance.
 A: 
let the area of $\triangle ABC$ be $1$ and $~\dfrac{AM}{MB}=\dfrac{\alpha}{1-\alpha}$ , $~\dfrac{BK}{KC}=\dfrac{\beta}{1-\beta}$ , $~\dfrac{CL}{LA}=\dfrac{\gamma}{1-\gamma}$    
$\triangle AML=\alpha(1-\gamma)$ , $~\triangle BKM=\beta(1-\alpha)$ , $~\triangle CLK=\gamma(1-\beta)$    
assume that $~\triangle AML>\dfrac{1}{4}$ , $~\triangle BKM>\dfrac{1}{4}$ , $~\triangle CLK>\dfrac{1}{4}$
$\Longrightarrow$ $\triangle AML\times \triangle BKM \times \triangle CLK>\dfrac{1}{64}$ ($\star$)   
also $~\triangle AML\times \triangle BKM \times \triangle CLK=\alpha(1-\alpha)\beta(1-\beta)\gamma(1-\gamma)\leq\dfrac{1}{64}$ ($\star$)
because $~\alpha(1-\alpha)\leq\dfrac{1}{4}$ , $~\beta(1-\beta)\leq\dfrac{1}{4}$ , $~\gamma(1-\gamma)\leq\dfrac{1}{4}~$ where $~0<\alpha$, $\beta$, $\gamma<1$    
thus, this is contradiction
and then at least one of them is less than or equal to a quarter of $~\triangle ABC$    
remark.
by coffeemath's advice, I add more detailed explanation. thanks.   
$\triangle AML=\dfrac{AM\cdot AL\cdot\sin A}{2}=\dfrac{\alpha AB\cdot (1-\gamma)AC\cdot \sin A}{2}$    
$=\alpha(1-\gamma)\dfrac{AB\cdot AC\cdot\sin A}{2}=\alpha(1-\gamma)\triangle ABC=\alpha(1-\gamma)$
A: Partial Sol, not sure if it leads anywhere
Since the Area is the half of product of two sides times sin of the angle between we get
$$\frac{\sigma AML}{\sigma ABC}=\frac{AM}{AB} \cdot \frac{AL}{AC}$$
$$\frac{\sigma BMK}{\sigma ABC}=\frac{BM}{AB} \cdot \frac{BK}{BC}$$
$$\frac{\sigma CKL}{\sigma ABC}=\frac{CK}{BC} \cdot \frac{CL}{CA}$$
Assuming by contradiction that all are greater than $\frac{1}{4}$ we get
$$AM \cdot AL \geq \frac{AB \cdot AC}{4}$$
$$BM \cdot BK \geq \frac{AB \cdot BC}{4}$$
$$CK \cdot CL \geq \frac{BC \cdot AC}{4}$$
Adding them together we get
$$AM \cdot AL+ BM \cdot BK+CK \cdot CL \geq \frac{AB \cdot AC+ AB \cdot BC +BC \cdot AC}{4}$$
Now, expressing $AB=AM+MB, BC=BK+KC, AC=CL+LA$ I expect we get a contradiction, but cannot see why the above inequality should be wrong...
Maybe someone can continue this....
A: Suppose $S(AML)>\frac{S(ABC)}{4},S(CKL)>\frac{S(ABC)}{4},S(MBK)>\frac{S(ABC)}{4}$
Now suppose one of the sides of $ABC$ is partitioned unequally.(If all sides are partitioned equally then the area of all smaller triangles will be $\frac{S(ABC)}{4}$).
For example, suppose $CK<KB$
$CK<KB \Rightarrow S(AKC)<\frac{S(ABC)}{2} \\ S(CKL)>\frac{S(ABC)}{4}\Rightarrow S(CKL)>\frac{S(AKC)}{2} \Rightarrow CL>AL$
$CL > AL \Rightarrow S(ABL)<\frac{S(ABC)}{2} \\ S(AML)>\frac{S(ABC)}{4} \Rightarrow S(AML)>\frac{S(ABL)}{2} \Rightarrow AM > BM$
$AM > BM \Rightarrow S(MBC)<\frac{S(ABC)}{2} \\ S(MBK)>\frac{S(ABC)}{4} \Rightarrow S(MBK)>\frac{S(MBC)}{2} \Rightarrow KB>CK$
This is a contradiction so the area of at least one of them must be less than or equal to $\frac{S(ABC)}{4}$
A: We may assume $B=(-1,0)$, $C=(1,0)$, $A=(0,\sqrt{3})$. Then $ABC$ is an equilateral triangle of side-length $2$. Let  $K=(t,0)$ with $-1<t<1$, and let $|MB|=x$ and $|LC|=y$. When the two triangles $BKM$ and $CKL$ both have an area at least ${1\over4}$ of the area of the  triangle $ABC$ then from the area formula involving two sides and the enclosed angle ($60^\circ$ in our case) it follows that necessarily
$$(1+t)x\geq1,\qquad(1-t)y\geq1\ ,$$
or
$$x\geq{1\over 1+t},\qquad y\geq{1\over1-t}\ .$$
This implies that
$$(2-x)(2-y) \leq{1+2t\over 1+t}\>{1-2t\over 1-t}={1-4t^2\over1-t^2}\leq1\ ,$$
and therefore
$${{\rm area}(AML)\over{\rm area}(ABC)}={(2-x)(2-y)\over4}\leq{1\over4}\ .$$
