= Two-point four-functions inequality
Suppose nonnegative numbers satisfy $a_0b_0\le c_0d_0$, $a_1b_1\le c_1d_1$, and $a_0b_1,a_1b_0\le c_1d_0$. Set $x=a_0b_1$, $y=a_1b_0$, $M=c_1d_0$, $N=c_0d_1$. Then $x,y\le M$ and $xy\le MN$. For $M>0$, $(M-x)(M-y)\ge0$ 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.
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