Two-point four-functions inequality

Suppose nonnegative numbers satisfy $a_0{b}_0\le c_0{d}_0$, $a_1{b}_1\le c_1{d}_1$, and $a_0{b}_1,a_1{b}_0\le c_1{d}_0$. Set $x=a_0{b}_1$, $y=a_1{b}_0$, $M=c_1{d}_0$, $N=c_0{d}_1$. Then $x,y\le M$ and $xy\le MN$. For $M>0$, $(M-x)(M-y)\ge 0$ gives $x+y\le M+xy/M\le M+N$; for $M=0$, $x=y=0$. Adding the two diagonal bounds proves $(a_0+a_1)(b_0+b_1)\le (c_0+c_1)(d_0+d_1)$. This lets the four functions theorem sum out one Boolean coordinate while preserving its pointwise hypothesis.