Bổ sung điều kiện $a,b,c$ dương
Ta có:
$\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge2$
$\to \dfrac{a}{a+1}\ge 2-\dfrac{b}{b+1}-\dfrac{c}{c+1}$
$\to \dfrac{a}{a+1}\ge \left(1-\dfrac{b}{b+1}\right)+\left(1-\dfrac{c}{c+1}\right)$
$\to \dfrac{a}{a+1}\ge \dfrac{1}{b+1}+\dfrac{1}{c+1}$
Áp dụng BĐT Cô-si, ta có:
$\dfrac{a}{a+1}\ge \dfrac{1}{b+1}+\dfrac{1}{c+1}\ge 2\sqrt{\dfrac1{(b+1)(c+1)}} \ \ (1)$
Hoàn toàn tương tự, ta có:
$\dfrac{b}{b+1}\ge 2\sqrt{\dfrac{1}{(a+1)(c+1)}} \ \ (2)$
$\dfrac{c}{c+1}\ge 2\sqrt{\dfrac{1}{(a+1)(b+1)}} \ \ (3)$
Nhân vế với vế của $(1), \ (2)$ và $(3)$, ta có:
$\dfrac{abc}{(a+1)(b+1)(c+1)}\ge 8\sqrt{\dfrac1{(a+1)^2(b+1)^2(c+1)^2}}=\dfrac{8}{(a+1)(b+1)(c+1)}$
$\to abc\ge8$
Dấu $=$ xảy ra $↔a=b=c=2$
