inequality involving sides and diagonals of quadrilateral

Question

Let $ABCD$ be a quadrilateral with $[ABCD]=1.\;$ Let $s$ be semi-perimeter; $\;p,q\;$ lengths of the diagonals, respecitively. Prove that: $$\;s+\frac{p+q}{2}\ge 2+\sqrt{2}.$$

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I tried to use inequality $s\ge\frac{p+q}{2}+z$, where $z$ is the distance between midpoints of diagonals.

Then it suffices to prove $p+q+z\ge 2+\sqrt{2}$, looks simpler because $s$ vanished, but $z$ appeared...

Answer 1

Let $AB=a$, $BC=b$, $CD=c$ and $DA=d$.

Hence, $$1=S_{ABCD}\leq S_{\Delta ABC}+S_{\Delta ACD}\leq \frac{1}{2}(ab+cd).$$

By the same way we'll obtain $1\leq\frac{1}{2}(ad+bc)$. Thus, $(a+c)(b+d)=ab+cd+ad+bc\geq4$.

In another hand, $1=S_{ABCD}\leq\frac{1}{2}pq$, which gives $pq\geq2$.

Id est, by AM-GM $$s+\frac{p+q}{2}=\frac{a+b+c+d}{2}+\frac{p+q}{2}\geq\sqrt{(a+c)(b+d)}+\sqrt{pq}\geq2+\sqrt2.$$ Done!