Let $ABCD$ be a convex quadrilateral and $k \gt 0$.
Let $M \in [AD]$, $N \in [BC]$ as $\frac{MA}{MD} = \frac{NB}{NC} = k$.
Show that if $AB = a$, $CD = c$, then $MN \le \frac{a+kc}{1+k}$.
I really don't know how to begin to solve this problem. Any advice or solution is really appreciated. Thanks!
This is one of those problems, where one needs to draw one or more auxiliary lines in order to visualize what the proof would look like. But first, we want to examine the inequality to be proven to see whether we can find an analogy. In other words, we are looking for a simpler case which manifests the same inequality. To make things easy, we consider a slightly weaker version of the given inequality, i.e. $$MN\lt \frac{a+kc}{1+k}.$$ We can express the same inequality in the following form. $$ a+kc \gt \left(1+k\right)MN \tag{1}$$
If you look closely, you will see that this resembles something familiar, i.e. the triangle inequality applied to a triangle with sidelengths $a$, $kc$, and $\left(1+k\right)MN$. Let us see whether we can construct such a triangle using the information given in your post.
We start the construction by drawing a line to join $M$ and the vertex $C$ (see Figure given above). Now, we need to draw a line parallel to $MN$ through the vertex $B$ and extend it to meet the line $CM$ at $P$. Finally, a line is drawn to join the vertex $A$ to $P$. Now, we are ready to get on with our proving.
By taking into account the similarity of the two triangles $BCP$ and $NCM$, we can write, $$\frac{PB}{MN}=\frac{CB}{CN}=1+k \qquad \mathrm{and}\qquad \frac{PM}{MC}=\frac{BN}{NC}=k.$$
The former equality gives us $$PB=\left(1+k\right)MN.$$ As a matter of fact, $PB$ is one of the side of the triangle we are seeking. The latter equality can be expressed as $$PM=kMC$$ and we are going to use this to find the side having the length $kc$. Consider the two triangles $AMP$ and $DMC$. It is given that $AM\space :\space MD=k\space :\space 1$. We just proved that $PM\space :\space MC=k\space :\space 1$. Furthermore, the two vertically opposite angles $\measuredangle AMP$ and $\measuredangle DMC$ are equal. Therefore, these two triangles are similar and we can state $$AP=kCD=kc.$$ So, $AP$ is another side of the triangle we are constructing. Since $AB=a$, $AB$ can be used as the third side.
The three line segments $PA$, $AB$, and $BP$ form the coveted triangle $ABP$. However, applying the triangle inequality to $ABP$ proves only the weak version of the inequality, i.e. inequality (1). Now, it is up to you to concoct a proof to show that the equality part of the original inequality holds, perhaps by considering certain special shapes of quadrilateral.