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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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I assume by symmetry you actually wanted the first term to be $a(1-b)$, right? – anon Sep 18 '11 at 8:40
One way to find the maximum of a nice expression like this on an interval is to take its gradient and set that to 0. The maximum must occur either on the boundary or where the gradient is 0. If it's on the boundary, then one of the variables is 0 or 1, and the expression simplifies and you can repeat taking the gradient in a lower dimensional space. – Logan Maingi Sep 18 '11 at 8:41
Furthermore, the LHS is at most $1-ab-cd$. – Did Sep 18 '11 at 9:46
@anon, yes, $2-ab-cd$. Thanks. – Did Sep 18 '11 at 10:02
up vote 5 down vote accepted

(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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I've LaTeXified your answer, hope you don't mind. And for the record (since my answer changed), my original trick involved seeing $$[a+(1-a)][b+(1-b)]=1,$$ which can be guessed at by seeing $a$ and $b$ as probabilities for independent events A and B and considering the probability of $$(A\text{ and }B)\text{ or }(A\text{ and not }B)\text{ or }(\text{not }A\text{ and }B)\text{ or }(\text{not }A\text{ and not }B).$$ – anon Sep 18 '11 at 9:43
@anon: I see now. Thanks. – valdo Sep 18 '11 at 11:46

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