Prove that square has the minimal area of all tangential quadrilaterals

I'm solving problems on extrema of multivariable functions from the Russian book by Vinogradova, Olehnik, Sadovnichy "Zadachi i uprazhnenija po matematicheskomu analizu", 2000. It's a problem on page 368, number 315:

"Prove, that square has the minimal area of all tangential quadrilaterals of a fixed circle"

I tried to use the formula for the area $S=\sqrt{abcd}\sin{\theta}$, where $a,b,c,d$ are sides and $\theta$ is the sum of two opposing angles. But when I equal partial derivatives by all five variables to zero, I don't get critical point $a=b=c=d$ What's wrong? I also don't think, that anything will be better if I use the condition $a+b-c-d=0$ and Lagrange multipliers. Maybe I should use another formula for the area and/or condition?

Let $r$ be a radius of the circle, $\alpha$, $\beta$, $\gamma$ and $\delta$ be measures of angles of our quadrilateral $ABCD$.
Thus, since $\cot$ is a convex function on $\left(0,\frac{\pi}{2}\right)$, by Jensen we obtain: $$S_{ABCD}=\frac{1}{2}(AB+BC+CD+DA)r=$$ $$=r^2\sum_{cyc}\cot\frac{\alpha}{2}\geq4r^2\cot\frac{\sum\limits_{cyc}\alpha}{8}=4r^2\cot45^{\circ}=4r^2.$$ The equality occurs for $\alpha=\beta=\gamma=\delta=90^{\circ}$
and from her $AB=AC=BC=CD=DA=2r,$ which says that it happens, when $ABCD$ is a square.
• Thank you, a nice solution. By the way, using your equality for the area (with cotangents), I added $\alpha+\beta+\gamma+\delta=2\pi$ condition and was able to solve it using Lagrange multipliers. Oct 19, 2017 at 16:35