Consider this nice little problem: if $ABCD$ is a quadrilateral inscribed in a unit square, then $$2\leq AB^2+BC^2+CD^2+DA^2\leq4$$ (Evidently this is problem 1 on paper 1 of the 1989 Irish Mathematical Olympiad. I also found it in another nice collection of problems, a recent admissions exam for the Indian Statistical Institute.)
Here's the straightforward algebraic proof I used: let $x_1$, $x_2$, $x_3$, $x_4$ denote legs of the right triangles with hypotenuses $AB$, $BC$, $CD$, $DA$, with no two of the $x_i$ lying on the same side of the square. Then the Pythagorean theorem gives $$AB^2+BC^2+CD^2+DA^2=\sum_{i=1}^4 x_i^2+(1-x_i)^2=\sum 1-2x_i(1-x_i)=4-2\sum x_i(1-x_i)$$ But $$0\leq x_i(1-x_i)\leq\frac{1}{4}$$ and the claim follows. (The lower bound of 2 is achieved when the vertices of $ABCD$ are the midpoints of the sides of the square, and the upper bound of 4 is achieved when $ABCD$ is the square itself.)
I'm curious about other approaches to the problem. Perhaps the result falls out of a more high-powered theorem in geometry? Or does the problem essentially boil down to a nice application of clever quadratic algebra?