Let $ABCD$ be a convex quadrilateral prove that an inequality holds true I've been trying to do the following problem:
Let $ABCD$ be a convex quadrilateral and points $E$ and $F$ on sides $AB,CD$ such that:
$\frac{AB}{AE}=\frac{CD}{DF}=n$
If $S$ is the area of $AEFD$ show that $\frac{AB*CD+n(n-1)AD^2+n^2*DA*BC}{2n^2}\ge S$

I attempted to solve it in the following way:
$\frac{AB*CD}{2n^2}=\frac{AE*DF}{2}$
$\frac{n^2*DA*BC}{2}=\frac{DA*BC}{2}$
$\frac{n(n-1)AD^2}{2n^2}=\frac{(n-1)AD^2}{2n}$
so we have that $\frac{AB*CD+n(n-1)AD^2+n^2*DA*BC}{2n^2}=\frac{AE*DF}{2}+\frac{DA*BC}{2}+\frac{(n-1)*AD^2}{2n}$
This is as far as I got, I couldn't think of any theorems or how to correlate these results with the area of $S$. Could you please continue my thought pattern and finish it off? Or explain why it can't be solved using my thought pattern and show an alternative approach?
 A: 
Although what you have started to do is correct, use of the given condition $\left(\mathrm{i.e.}\space \frac{AB}{AE}=\frac{DC}{DF}=n\right)$ alone is not going to lead us to the required proof. You have to introduce the area of the quadrilateral $S$ into the picture by bringing certain theorems, inequalities and other observations into play. We start by proving a simple lemma, which we need later. We also assume that $\infty\ge n\ge 1$.
$\underline{\mathrm{1.\space Lemma}}$
$TUVW$ is a convex quadrilateral (see $\mathrm{Fig.\space 1.1}$). Points $X$ and $Y$ divide its sides $TU$ and $VW$ respectively, such that $\frac{TU}{TX}=\frac{WV}{WY}=n$. The following inequality is true for all $n$, while the equality holds only when the two sides $UV$ and $WT$ are parallel to each other.
$$\left(\frac{n-1}{n}\right)WT+\left(\frac{1}{n}\right)UV\ge XY \tag{1}$$
$\underline{\mathrm{1.1.\space Proof}}$
Draw a line parallel to $WT$ through $X$ to meet the diagonal $UW$ at $Z$. This makes $\frac{WU}{WZ}=n$. Now, draw another line, this time parallel to $UV$, through $Y$ to meet the diagonal $UW$ at some point $Z_1$. This gives us $\frac{WU}{WZ_1}=n$. Since $\frac{WU}{WZ_1} = \frac{WU}{WZ}$, the two points $Z_1$ and $Z$ are one and the same.
When we apply the triangle inequality to the triangle $XYZ$, we have $ZX+YZ\gt XY$. If $WT$ is parallel to $UV$, it is obvious that $Z$ lies on $XY$. Therefore, only in this particular instance, we can write $ZX+YZ= XY$. Therefore, in general, it is true that
$$ZX+YZ\ge XY. \tag{2}$$
We know that $ZX= \left(\frac{n-1}{n}\right)WT$ and $YZ=\left(\frac{1}{n}\right)UV$. When we substitute these values in inequality (2), we get,
$$\left(\frac{n-1}{n}\right)WT+\left(\frac{1}{n}\right)UV\ge XY.$$
$\underline{\mathrm{2.\space Ptolemy’s\space Theorem\space and\space Inequality}}$
Consider $\mathrm{Fig.\space 1.2}$. Ptolemy’s theorem states that, for the cyclic quadrilateral shown in that diagram, the sum of products of the two pairs of opposite sides equals the product of its diagonals, i.e. $AC\times BD = AB\times CD + BC\times DA$. However, for a quadrilateral which is not cyclic, Ptolemy’s theorem becomes an inequality, i.e. $AC\times BD \lt AB\times CD + BC\times DA$. Therefore, in general, it is true that,
$$AC\times BD \le AB\times CD + BC\times DA. \tag{3}$$
$\underline{\mathrm{3.\space An\space Area\space Inequality\space for\space Convex\space Quadrilaterals}}$
Consider $\mathrm{Fig.\space 1.3}$. The area $S$ of the quadrilateral shown in that diagram can be expressed as $2S= AC\times BD\sin\left(\phi\right)$, where $\phi$ is the angle between its two diagonals $AC$ and $BD$. For a given pair of diagonals, the left-hand side of this expression has its maximum value, if they are perpendicular to each other, i.e. $\phi=90^o$. Therefore, we can write,
$$S\le \frac{AC\times BD}{2}. \tag{4}$$
$\underline{\mathrm{4.\space Proof\space of\space the\space Inequality}}$

Now, we have all the tools we need to prove the given inequality. Please pay your attention to $\mathrm{Fig.\space 2}$ and also note that we start off by proving a slightly different inequality, i.e.
$$\frac{AB\times CD+n\left(n-1\right)AD^2+nDA\times BC}{2n^2}≥S, \tag{5}$$
in which the last term of the numerator on the left-hand side is not $\color{red}{n^2}DA\times BC$, but $\color{red}{n}DA\times BC$.
First of all, we expand the left-hand side of inequality (5) to to get,
$$\frac{AB\times CD+n\left(n-1\right)AD^2+nDA\times BC}{2n^2}\qquad\qquad =\frac{1}{2}\left[\frac{AB}{n}\times\frac{CD}{n}+AD\left\{\left(\frac{n-1}{n}\right)AD+\frac{1}{n}BC\right\}\right].$$
As exactly as you have done, we too proceed by eliminating $AB$ and $DC$ using the given relations $AB=nAE$ and $CD=nDF$.
$$\frac{AB\times CD+n\left(n-1\right)AD^2+nDA\times BC}{2n^2}\space\quad\qquad =\frac{1}{2}\left[AE\times DF+AD\left\{\left(\frac{n-1}{n}\right)AD+\frac{1}{n}BC\right\}\right]$$
Now, we use the lemma we have proven above to obtain the following inequality.
$$\frac{AB\times CD+n\left(n-1\right)AD^2+nDA\times BC}{2n^2}≥\frac{1}{2}\left(AE\times DF+AD\times EF\right)$$
Next, with the help of Ptolemy’s inequality, we simplify the right-hand side of the above inequality as shown below.
$$\frac{AB\times CD+n\left(n-1\right)AD^2+nDA\times BC}{2n^2}≥\frac{1}{2}AF\times DE$$
Finally, we use the area inequality for quadrilaterals to complete our proof of inequality (5), which is valid for all $\infty\ge n\ge 1$.
$$\frac{AB\times CD+n\left(n-1\right)AD^2+nDA\times BC}{2n^2}≥S $$
As shown below, we can now multiply the last term of the numerator on the left-hand side of inequality (5) by $n$ to obtain the inequality you have stated in your question, because $n\ge 1$, i.e.
$$\frac{AB\times CD+n\left(n-1\right)AD^2+n^2DA\times BC}{2n^2}≥S. \tag{6}$$
Inequality (6) still preserves its validity, however, it becomes weaker than inequality (5).
