Prove that at least one area is less then one quarter of area of ABC Question- 
$P,Q,R$ are points on the sides $BC,CA,AB$ of $\Delta ABC$. Prove that the area of at least one of the triangles $AQR, BRP, CPQ$ is less than or equal to one quarter of the area of $\Delta ABC$.
My try - 
Now what i try is that if there are 4 sub triangles (+ PRQ) in the question then I can directly apply piegon hole principle average form to get result...but I have no idea how to prove these for 3 triangles given in question...
Any hint. Will be of great importance....
Thankyou
Source - CTPCM (Olympiad book)
Figure :
(https://i.stack.imgur.com/DjbeU.jpg)
 A: In the standard notation let $BP=xa,$ $CP=(1-x)a,$ $CQ=yb$, $AQ=(1-y)b$, $AR=zc$ and $RB=(1-z)c,$ where $\{x,y,z\}\subset(0,1).$
Thus, by AM-GM $$\prod_{cyc}\frac{S_{\Delta ARQ}}{S_{\Delta ABC}}=\prod_{cyc}z(1-y)=\prod_{cyc}x(1-x)\leq\prod_{cyc}\left(\frac{x+1-x}{2}\right)^2=\frac{1}{64},$$
which says that one of numbers in the first product is less or equal to $\frac{1}{4},$ 
otherwise we'll got a contradiction. 
A: Using barycentric coordinates :
$$P=[0,p,1-p], \ \ \ Q=[1-q,0,q], \ \ \ R=[r,1-r,0]$$
with, of course :
$$A[1,0,0], \ \ \  B[0,1,0], \ \ \ C[0,0,1].$$
The ratio of area of triangle (ARQ) to area of triangle (ABC) is $\det([A;R;Q])=q(1-r)$
The ratio of area of triangle (BPR) to area of triangle (ABC) is $\det([B;P;R])=r(1-p)$
The ratio of area of triangle (CQP) to area of triangle (ABC) is $\det([C;Q;P])=p(1-q)$
Let us assume the contrary, i.e.,
$$\begin{cases}q(1-r)>\tfrac14\\
r(1-p)>\tfrac14\\
p(1-q)>\tfrac14\end{cases}$$
Taking the product (all quantities are $\geq 0$) :
$$(p(1-p))(q(1-q))(r(1-r))>\tfrac{1}{64}\tag{1}$$
But for any $x$ with $0 \leq x \leq 1$, we have $x(1-x)\leq \dfrac14$.
Therefore (1) cannot hold : we have established a contradiction.
