inequality involving the product of lengths of edges of quadrilateral Is there a constant $C$ such that for every quadrilateral with sides $a,b,c,d$ and area $S$, the following inequality holds:
$$abcd\ge C\cdot S^2$$

My attempt:
It is known, that $$S^2\le (s-a)(s-b)(s-c)(s-d)=\frac{1}{16}(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)$$ with possible equality, so we have to have
$${abcd}\ge\frac{C}{16}(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)$$
Of course $C\le 1$ (take $a=b=c=d$). Let's try to prove that $C=1$ works fine, since I found no counterexample.
The inequality we want to prove is equivalent to
$$a^4+b^4+c^4+d^4+8abcd\ge 2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)$$
and I stucked.

There is no positive constant $C$. Take a non degenerate convex quadrilateral $ABCD$  and then move $D$ to $A$ along the initial side $DA$. Then the product $abcd$ goes to zero (because $d=|DA|$ goes to zero and $c=|DC|$ remains bounded). Moreover  $S$ is greater or equal to the area of the triangle $ABC$ which is positive and does not change.
$$0\leftarrow abcd\geq C\cdot S^2\geq C|\triangle ABC|^2>0$$
