#### Solution

The main idea of this solution is to replace the inside quadrilateral with a series of figures, such that each next one has a larger perimeter, and to eventually reach the outside quadrilateral. That will prove that the perimeter of $ABCD$ is smaller than that of $EFGH$. $\\$

Let’s proceed in a similar manner as in the Example 3. First, extend any of the sides of the inside quadrilateral such that it touches a side of the outside one.

Then, by triangle inequality, we can write that:

$$CI + IB > BC$$

So the perimeter of $AICD$ is larger than the perimeter of $ABCD$. We repeat this until all the vertices are on sides of the outside quadrilateral. There are actually a couple of different cases here, but they are all very similar to each other. We can have one vertex per side, or we can have one or two sides that have two vertices on them.

In essence, the next step boils down to showing that the distance between any two consecutive vertices of the new quadrilateral $IJKL$ is the shortest along the side connecting them, and in particular, shorter then if we were to go around the perimeter of $EFGH$. Which is true, because a straight segment is the shortest path connecting two points. $\\$

In the particular case above, we can divide that into steps, for example first show that $IH + HL > IL$, extending the inner quadrilateral towards one of the vertices of the outer one. Then we can cut the corners, by for example saying that $JF + IF > IJ$. In any way, we show that the perimeter of $EFGH$ is larger than the perimeeter of $IJKL$, which is larger than the perimeter of $ABCD$ and we reach our objective here.