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