Solution: Name the 15 sides $a_{1}, \ldots, a_{15}$ and define $a_{16}=a_{1}$. Consider the sums $b_{i}=$ $a_{i}+a_{i+1}$ for $i=1, \ldots, 15$. Dann gilt $b_{1}+\ldots+b_{15}=2\left(a_{1}+\ldots+a_{15}\right)=42$. Using the pusher compartment principle, it follows that there is a $b_{i} \leq \frac{42}{15}=\frac{14}{5}$ exists. The triangle, which has the two side lengths $a_{i}$ and $a_{i+1}=b_{i}-a_{i}$, can therefore span at most the area $\frac{1}{2} a_{i}\left(b_{i}-a_{i}\right)$, i.e. at most $\frac{1}{2} a_{i}\left(\frac{14}{5}-a_{i}\right)$. We therefore want to show that $\frac{1}{2} a_{i}\left(\frac{14}{5}-a_{i}\right)\frac{7}{5}$ follow, so we are done.