<s>[INST] show that inequality holds for real numbers 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 
$b(1-a)+c(1-a)+d(1-a)-(1-a)\le b(c+d)+cd$ 
$(1-a)(b+c+d-1) \le b(c+d)+cd$ 
 [/INST] 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
$$
   a+b+c+d+3abcd \le 1+a(b+c+d)+b(c+d)+cd
$$
</s>