There are given real numbers $a,b,c,d \in [0,1]$ show that:

$a+b+c+d \le 1+a(b+c+d)+b(c+d)+cd$

I tried to transform it to

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

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


Expanding $0 \le (1-a)(1-b)(1-c)(1-d)$ gives

$$ 0 \le 1-a-b-c-d+ab+ac+ad+bc+bd+bd - abc-abd-acd-bcd+abcd $$

Each of the terms $abc$, $abd$, $acd$, $bcd$ is $\ge abcd$, so it follows that

$$ 0 \le 1-a-b-c-d+ab+ac+ad+bc+bd+bd - 3 abcd $$

which gives the slightly stronger statement

$$ a+b+c+d+3abcd \le 1+a(b+c+d)+b(c+d)+cd $$

  • $\begingroup$ Hey, could you help me with the general version of the inequality which is $\displaystyle \sum_{1\le i \le n } a_i \le 1 +\sum_{1 \le i}\sum_{<j\le n} a_ia_j$ I'm not sure if I can use your idea with suplemental inequality $\endgroup$ – Gregor Apr 24 '14 at 11:00

Note that your inequality is linear in each of its variables. A consequence of this is that it suffices to verify it at the $2^{4}$ corners of the hypercube (this follows since, if you fix any 3 of the variables, the resulting expression will be a linear expression in one variable which will be maximized at either 0 or 1).


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