a, $\frac{1}{a+1}=1-\frac{1}{b+1}+1-\frac{1}{c+1}+1-\frac{1}{d+1}$
$<=> \frac{1}{a+1}=\frac{b}{b+1}+\frac{c}{c+1}+\frac{d}{d+1}\geq 3\sqrt[3]{\frac{bcd}{(b+1)(c+1)(d+1)}}$
Tương tự có 3 cái còn lại rồi nhân vế theo vế là oke
b,$P= \frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}$
$<=>P+4=\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}=(a+c)(\frac{1}{b+c}+\frac{1}{d+a})+(b+d)(\frac{1}{c+d}+\frac{1}{a+b})\geq (a+c)\frac{4}{a+b+c+d}+(b+d)\frac{4}{a+b+c+d}=4$
$=> p \geq 0$