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 $D$. 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 $D''$.

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 $D''$.

It seems that I found a proof. Consider a figure $F$ with minimal ratio $S/d^2$.

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 $D$. 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 $D''$.

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 $D$. 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 $D''$.