Toán 9 Bất đẳng thức

$$a+b+c\ge \frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}=\left(a+1-\frac{b(a+1)}{b+1}\right)+\left(b+1-\frac{c(b+1)}{c+1}\right)+\left(c+1-\frac{a(c+1)}{a+1}\right)=a+b+c+3-\frac{b(a+1)}{b+1}-\frac{c(b+1)}{c+1}-\frac{a(c+1)}{a+1}$$
$$\Leftrightarrow \frac{b(a+1)}{b+1}+\frac{c(b+1)}{c+1}+\frac{a(c+1)}{a+1}\ge 3$$
Áp dụng BĐT AM-GM $$\frac{b(a+1)}{b+1}+\frac{c(b+1)}{c+1}+\frac{a(c+1)}{a+1}\ge 3\sqrt[3]{\frac{b(a+1)}{b+1}\cdot \frac{c(b+1)}{c+1}\cdot \frac{a(c+1)}{a+1}}=3\sqrt[3]{abc}\ge 3$$