Let $H$ be a compact, convex subset of $\Bbb{R}^2.$ For a given $m\ge 3,$ let $P_m$ be a polygon of maximum area which contained in $H$ and has atmost $m$ sides. Then $$\dfrac{Area(P_m)}{Area(H)}\ge\dfrac{m}{2\pi}\sin\left(\dfrac{2\pi}{m}\right)$$ and equality holds if and only if $H$ is an ellipse.
I found this result fom here and has no clue to prove this. Any idea or a (easy) reference to attempt this problem?
I think even a bound $\ge \epsilon_m$ has some value. I will provide an $\epsilon_3$.
Consider $H$ compact convex body. Take $\Delta A_1 A_2 A_3$ of largest area. The points $A_i$ are on the boundary. Because we cannot enlarge the area by moving $A_1$, there exists a supporting line to $H$ through $A_1$ that is parallel to $A_2 A_3$. Similarly for the other vertices. Thus we can inscribe $H$ in a larger triangle $\Delta A_1'A_2'A_3'$ for which the $A_i$'s are the midpoints. Therefore $\operatorname{Area}\Delta A_1 A_2 A_3 \ge \frac{1}{4}\operatorname{Area}H$.
Notice that if $H$ were the ellipse tangent to the sides of the larger triangle at the midpoints $A_i$'s then we would have the exact bound stated ( the figure would be affinely equivalent to a circle inscribed in a equilateral triangle).