The following "piecewise-quadratic" inequality emerged in a joint work of Rom Pinchasi and myself. The inequality is surprisingly delicate, and all our attempts to simplify it made it false. By the end of the day, we were able to prove the inequality, but the proof is unreasonably sophisticated, totalling to about 15 pages. We would be happy to have a shorter proof.
The inequality involves the function $G$ of three real variables, defined as follows: if $(\xi,\eta,\zeta)$ is a non-decreasing rearrangement of $(x,y,z)$, then we let $$ G(x,y,z) := \begin{cases} \xi\eta &\ \text{if}\ \zeta\ge \xi+\eta, \\ \xi\eta-\frac14\,(\xi+\eta-\zeta)^2 &\ \text{if}\ \zeta\le \xi+\eta. \end{cases} $$ Thus, for instance, we have $G(9,6,7)=38$, whereas $G(7,14,6)=42$. Now consider the function $f$ of four variables defined by $$\begin{align*} f(x_0,x_1,y_0,y_1) &:= \min \{ 0.15s^2, x_0y_0+x_1y_1 \} \\ &\qquad + G(x_0,y_1,1-s) + G(x_1,y_0,1-s) \\ &\qquad + 0.25(1-s)^2, \end{align*}$$ where for brevity I write $s=x_0+x_1+y_0+y_1$, and let $$ \Omega := \{ (x_0,x_1,y_0,y_1)\in{\mathbb R}_{\ge 0}^4\colon 1/2 \le s \le 1. \}. $$ All we want to show is that $$ \max_\Omega f \le 0.15. $$ (Indeed, the maximum is actually equal to $0.15$: say, we have $f(0.5,0,0.5,0)=0.15$.)
At Pat Devlin's suggestion, here is the graph of the maximum as a function of the sum $s=x_0+x_1+y_0+y_1$.
(source: haifa.ac.il)
It looks nice, but does not seem to be a graph of some "simple" function; and so, there is probably no simple analytic expression for $\max f$ over all quadruples $(x_0,x_1,y_0,y_1)$ adding up to $s$.