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Given any $a, b, c, d \in [0, 1],$ how can the following be proven?

$a (1-b) + b(1-a) + c(1-d) + d(1-c) \le 2$

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    @anon, yes, $2-ab-cd$. Thanks.2011-09-18

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(To those who originally upvoted: I've changed my answer because this is a much sleeker version, not sure why I didn't see it before.)

This is just a single inequality added to itself with $c,d$ instead of $a,b$ in the second copy. Namely, $a(1-b)+(1-a)b\le a(1)+(1-a)(1)\le1.$

Above we used the fact that if $b\in[0,1]$, then $b\le1$ and $1-b\le1$.

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I agree with anon's solution. I'd also like to post mine. Anon's solution is based on a "trick" with inequalities, which is understood, but not so simple to guess from the beginning. OTOH my solution is straightforward.

$S = a(1−b)+(1−a)b$

Let's find the derivative of the above with respect to $a$.

$\frac{dS}{da} = 1-2b.$

From the above we see that the derivative of $S$ with respect to $a$ is constant (i.e. does not depend on $a$). Hence the extremal values are achieved at the ends of the range.

  • Assuming $1-2b > 0, b < 1/2$. The maximum is achieved when $a=1$. Substituting this we get $S = 1-b$. Since $b\ge0$ we get $S \le 1$.
  • Assuming $1-2b < 0, b > 1/2$. The maximum is achieved when $a=0$. Substituting this we get $S = b$. Since $b<=1$ we get $S \le 1$.
  • If $1-2b = 0$, then $b = 1/2, S = 1/2 \le 1$.
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    @anon: I see now. Thanks.2011-09-18