How prove this $S_{\Delta ABC}\ge\frac{3\sqrt{3}}{4\pi}$ 
There is convex body $T$ (with the area is $1$), show that
  there is a triangle $\Delta ABC$, such $A,B,C\in T$, and 
  $$S_{\Delta ABC}\ge\dfrac{3\sqrt{3}}{4\pi}$$


This problem is from China The Olympic book, but it doesn't show a solution in book, it says it is a "W.Blaschke 1917" result, but I can't find it.
Can you give me a reference or solve it?
 A: Yesterday I went to the ETH Zurich library and got hold of Wilhelm Blaschke's 1917 paper 
Über affine Geometrie III: Eine Minimumeigenschaft der Ellipse. Leipziger Berichte 69 (1917), pages 3–12. 
In this paper he proves that any convex body $B$ in the plane contains a triangle $T$ with
$${\rm area}(T)\geq{3\sqrt{3}\over 4\pi}{\rm area}(B)\ .$$
The proof uses so-called Steiner symmetrization. One symmetrization step (in $y$-direction) consists in replacing
$$B:=\bigl\{(x,y)\>\bigm|\>a\leq x\leq b, \quad c(x)\leq y\leq d(x)\bigr\}$$
by
$$B':=\bigl\{(x,y)\>\bigm|\>a\leq x\leq b, \quad c(x)-m(x)\leq y\leq d(x)-m(x)\bigr\}\ ,$$
where
$$m(x):={c(x)+d(x)\over2}\ ;$$
see the figure below, taken from Blaschke's paper. 

By Cavalieri's principle we can conclude that
$${\rm area}(B')={\rm area}(B)\ .$$Denote the area of a maximal triangle in $B$ and $B'$ by $\Delta$ and $\Delta'$ respectively. Draw in $B'$ two mirror copies $T'$, $\bar T'$ of maximal triangles, and let $T$, $\bar T$ be their inverse images in $B$. Then one shows with a simple computation that
$$2\Delta\geq{\rm area}(T)+{\rm area}(\bar T)={\rm area}(T')+{\rm area}(\bar T')=2\Delta'\ .$$
Symmetrizing repeatedly in different directions we obtain a sequence of convex bodies $(B_n)_{n\geq0}$. All $B_n$ have the same area, and the areas of their maximal triangles are monotonically decreasing. By choosing a suitable  sequence of directions we can make sure that the $B_n$ converge to a circular disk $D$ having the same area as $B_0=B$. (For the proof of this Blaschke refers to his monograph Kreis und Kugel, Leipzig 1916). It follows that
$$\Delta(B)\geq \Delta(D)={3\sqrt{3}\over 4\pi}{\rm area}(D)={3\sqrt{3}\over 4\pi}{\rm area}(B)\ .$$
A: I can't add comments since my reputation, so I use this to comment.
I just list something easy to notice about:
If $T$ is the circle, then the triangle with the maximal area in it is the equilateral triangle, so the area of it is exactly $\frac{3\sqrt 3}{4\pi}$. Thus if $T$ is the ellipse with the area equals to $1$, it is similarly to the case of the circle.
If $T$ is the equilateral triangle, then obviously the triangle with the maximal area in it is itself and $1\geq \frac{3\sqrt 3}{4\pi}$.
