Suppose $LD=y,AL=x$ now so $[LKD]=\frac {x^2}{(x+y)^2}[ABD],[ALM]=\frac {x^2}{(x+y)^2}[ACD]$. So $[LKM]=\frac {x^2[ACD]-xy[ABD]}{(x+y)^2}$. Now as $\frac {LD}{AL}=\frac {MC}{AM}$ so $LK=MN$ and that implies $MNKL$ is nothing but a parralelogram.Now take $[AKD]=py,[AKB]=px,[BKC]=xt\implies [CKD]=yt$.Now so certainly $[ABCD]=(p+t)(x+y)$. And also $[KLMN]=2(x^2y(p+t)-xy(x+y)p$. So finally now we're required to $\frac {xy}{(x+y)^3}(\frac {xt-yp}{p+t})<\frac {4}{27}$.Now we can take $t\geq p$. So now keep $x,y,p$ constant, then $f(t)=(\frac {xt-yp}{p+t})$ is increasing. So $f(t)\leq \lim_{t\to\inf} f(t)= x$.Thus it's now remain to show $\frac {x^2y}{(x+y)^3}<\frac{4}{27}$. Which is obvious just by $AM-GM$ inequality,So done.

I dont see how the AM-GM gives you that result.

NgoNgang wrote: I dont see how the AM-GM gives you that result. Just use on $\frac {x}{2},\frac{x}{2},y$

Another way is to use $ { \frac {x}{x + y}, \frac {x}{x + y}, \frac {2y}{x + y} }$

$ \frac{x^2y}{(x+y)^3}<= \frac{4}{27} $ and not < (smaller). There is a positive element that is greater than zero on the denominator to make it smaller than zero.