, 2010

Without loss of generality we can assume that $a$, $b$, $c$ and $d$ are the lengths of consecutive sides of the quadrilateral. A diagonal divides the quadrilateral into two triangles. From one partition we get the inequality $\frac{ab}{2} + \frac{cd}{2} \ge S$, whence $ab + cd \ge 2S$, and from the other partition $\frac{bc}{2} + \frac{da}{2} \ge S$, whence $bc + da \ge 2S$. Therefore $ab + bc + cd + da \ge 4S$.

On the other hand, by adding the inequalities $a^2 + b^2 \ge 2ab$, $b^2 + c^2 \ge 2bc$, $c^2 + d^2 \ge 2cd$, and $d^2 + a^2 \ge 2da$, and dividing by 2 we get $a^2 + b^2 + c^2 + d^2 \ge ab + bc + cd + da$, which implies the required inequality.

The equality holds iff all the inequalities used are, in fact, equalities. In the first inequality the equality holds iff all the angles are right angles. In the second step the equalities hold iff all sides are of equal length.