Geometry Inequality

[RoSe-JoKer]

Let $Q$ be any convex quadrilateral of area $F$ and semiperimeter $s$.
Suppose that
length of any diagonal of $Q$ $\geq$ length of any side of $Q$ $\geq 1$
Prove that
$\frac {2}{\sqrt3}F \geq s - 1$

[Anton]

Problem. Let $Q$ be a convex quadrilateral of area $F$ and semiperimeter $s$. Suppose that the length of any diagonal of $Q$ is greater than or equal to the length of any side of $Q$, and that the length of any side of $Q$ is greater than or equal to a given length $\ell>0$. Prove that
$$F\geq \frac{\sqrt{3}}{2}\,\ell\,(s-\ell)\,.$$

Let $ABCD$ denote such a quadrilateral. We shall first prove that every internal angle of $ABCD$ is at least $\dfrac{\pi}{3}$. To show this, we note that
$$AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos(\angle ABC)=AC^2\geq \max\{AB^2,BC^2\}\,.$$
Therefore,
$$\cos(\angle ABC)\leq \frac12\,\min\left\{\frac{AB}{BC},\frac{BC}{AB}\right\}\leq \frac12\,.$$
Thus, $\angle ABC\geq \dfrac{\pi}{3}$. Similarly, $\angle BCD$, $\angle CDA$, and $\angle DAB$ are also at least $\dfrac{\pi}{3}$.

Let $a:=AB$, $b:=BC$, $c:=CD$, and $d:=DA$. Because the sum of the internal angles of a quadrilateral is $2\pi$, there exists a pair of opposite angles of the quadrilateral whose sum is at most $\pi$. Without loss of generality, suppose that $\angle ABC+\angle CDA\leq \pi$. Because $\angle ABC\geq \dfrac{\pi}{3}$, we obtain $\angle CDA\leq \dfrac{2\pi}{3}$. Similarly, since $\angle CDA\geq \dfrac{\pi}{3}$, we obtain $\angle ABC\leq \dfrac{2\pi}{3}$. Consequently,
$$\sin(\angle ABC)\geq \frac{\sqrt{3}}{2}\text{ and }\sin(\angle CDA)\geq \frac{\sqrt{3}}{2}\,.$$
Thus,
$$F=\frac{1}{2}\,a\,b\,\sin(\angle ABC)+\frac{1}{2}\,c\,d\,\sin(CDA)\geq \frac{\sqrt{3}}{4}\,(ab+cd)\,.$$
Since $a$, $b$, $c$, and $d$ are greater than or equal to $\ell$, we have
$$(a-\ell)(b-\ell)\geq 0\,,$$
so
$$\frac{ab}{2}\geq \ell\left(\frac{a+b}{2}-\frac{\ell}{2}\right)\,.$$
Likewise,
$$\frac{cd}{2}\geq \ell\left(\frac{c+d}{2}-\frac{\ell}{2}\right)\,.$$
Thus,
$$F\geq \frac{\sqrt{3}}{2}\,\left(\frac{ab+cd}{2}\right)\geq \frac{\sqrt{3}}{2}\,\left(\ell\left(\frac{a+b}{2}-\frac{\ell}{2}\right)^\vphantom{2^2}+\ell\left(\frac{c+d}{2}-\frac{\ell}{2}\right)\right)\,.$$
That is,
$$F\geq \frac{\sqrt{3}}{2}\,\ell\,\left(\frac{a+b+c+d}{2}-\ell\right)=\frac{\sqrt{3}}{2}\,\ell\,(s-\ell)\,.$$
Note that the equality holds if and only if $s=2\ell$ and $Q$ is a diamond rhombus with side length $\ell$ (i.e., two opposite angles of $Q$ are of measure $\dfrac{\pi}{3}$, and the other two angles are of measure $\dfrac{2\pi}{3}$).