Proof of an inequality

  • #1
stevendaryl
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I'm pretty sure that the following is true, but I don't see an immediate compelling proof, so I'm going to throw it out as a challenge:

Let [itex]A,A', B, B'[/itex] be four real numbers, each in the range [itex][0,1][/itex]. Show that:

[itex]AB + AB' + A'B \leq A' B' + A + B[/itex]

(or show a counter-example, if it's not true)

This inequality was inspired by Bell's Theorem, but that's not relevant to proving or disproving it.
 

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  • #2
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Consider the region ##A+A' \leq 1## first.
For B=0, the inequality simplifies to ##AB' \leq A'B'+A## which is true.
The derivatives with respect to B are ##A+A'## and ##1##, respectively, which means the derivative for the left hand side is smaller or equal. For the chosen region this means the inequality stays true for all B.

Now consider the region ##A+A' > 1##.
For B=1, the inequality simplifies to ##AB'+A' \leq A'B'+1##. As ##B'+A' \leq A'B'+1##, this inequality is satisfied.
The derivatives with respect to B are ##A+A'## and ##1##, respectively, for the chosen region the derivative on the left hand side is larger, but now we are going backwards with B. For the chosen region this means the inequality stays true for all B.


For ##A=1-A'##, the inequality can be written as ##AB+AB'+(1-A)B \leq (1-A)B'+A+B## which simplifies to ##2AB' \leq B'+A##, that inequality does not depend on B and it is true with enough margin to make the derivatives work on both sides.
 
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  • #3
stevendaryl
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Thanks! That's a little more elegant than the only way I found to prove it, which was to do an exhaustive case split.
 

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