Diameter-area ratio for affine tranformations. Consider an convex plane figure $F$. How to prove that there is an affine transformation $a$ such that $\sqrt{3}$ diameter$(a(F))^2\leq 4$ area$(a(F))$?
I found only one reference, to "Über einige Affininvarianten konvexer Bereiche", but unfortunately it is in German.
Added: formula (12) there looks like desirable.
After I found a solution myself, I can understand German. The proof there in the pages 734 (corresponding to considering $D'$ below) and 735 (considering $D''$).
The author estimated $f/d_u^2$, $f$ is an area (Flacheninhalt) and $d_u$ is a diameter(Durchmesser).
So, emergency over, thank you))
the proof is rewritten by me in http://arxiv.org/abs/1306.4688
 A: The article of Scott and Awyong, "INEQUALITIES FOR CONVEX SETS" containes several inequalities
relating the area, perimeter, width, diameter, inradius and circumradius of planar convex sets.
I tried to obtain the desired inequality (all of them have precise references given). Denote by $A$ the area,
and by $d$ the diameter. At least, in the case $p\ge 3d$ (and I hope this will help) we obtain
$$
A \ge \frac{\sqrt{3}}{4}d(p-2d)\ge \frac{\sqrt{3}}{4} d^2.
$$
I did not check if we can obtain the inequality in another way. Some references claim that the
inequality has been proved by Behrend (the german article), others relate it to Kubota.
A: It seems that I found a proof. Consider a figure $A$.
Consider a figure $F$ with minimal ratio $S/d^2$ among all affine transforms of $A$.
Lemma. There are two diameters of $F$ with angle at least $\pi/3$ between them.
Note that Lemma implies the estimation because $\sin(\pi/3) =\sqrt 3/2$ and area of $F$ is at least $d^2\sin(\pi/3)/2$.
Proof.
Consider a diameter $D$. Try to squeeze $F$ in the direction of $D$ and stretch out in the perpendicular direction.
It is not possible, therefore there is an other diameter $D'$ with angle at least $\pi/4$ and less than $\pi/3$ with $D$. 
Well, among all pairs of diameters chose the pair $D,D'$ with the biggest shapr angle between them.
Now try to perform an affine shift parallel to $D$ in the direction decreasing $D'$.
It is not possible, therefore there is a diameter $D''$ which is a kind in a "symmetric" position with $D'$. So, now either the angle between $D$ and $D'$ is big enough, or the angle between $D'$ and $D''$. 
