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Suppose $a,b,c,d$ are real numbers greater than $1$. Prove that $$abc+abd+acd+bcd-3abcd<1.$$

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I have downvoted this post because there is no context for the problem, work shown towards its solution, or anything at all besides the statement of the problem. I'd be happy to revert this if you remedy this in an edit. –  mixedmath Jun 5 '13 at 18:43

2 Answers 2

up vote 10 down vote accepted

Let $a=1+x$, $ b=1+y$, $ c=1+z$ and $d=1+w$. Then $abc+acd+abd+bcd-3abcd=1-(xy+xz+xw+yz+yw+zw)-2(xyz+xzw+xyw+yzw)-3xyzw$

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Very nice ... +1 –  user1551 Jun 5 '13 at 21:06

The inequality holds because \begin{align*} &4\times\left[\,3abcd-(abc+abd+acd+bcd)+1\,\right]\\ \\ =&2(a-1)(b-1)(c-1)(d-1)\\ \\ &+(a+1)(b+1)(c-1)(d-1)+(a+1)(b-1)(c+1)(d-1)\\ &+(a+1)(b-1)(c-1)(d+1)+(a-1)(b+1)(c+1)(d-1)\\ &+(a-1)(b+1)(c-1)(d+1)+(a-1)(b-1)(c+1)(d+1)\\ \\ &+(a+1)(b-1)(c-1)(d-1)\\ &+(a-1)(b+1)(c-1)(d-1)\\ &+(a-1)(b-1)(c+1)(d-1)\\ &+(a-1)(b-1)(c-1)(d+1). \end{align*}

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