Đề: Cho $a,b,c>0$ thõa mãn $ab\ge 1; \ c(a+b+c)\ge 3$. Tìm Min:
$$P=\dfrac{b+2c}{1+a}+\dfrac{a+2c}{1+b}+6\ln(a+b+2c)$$

Bài Làm:
$$P=\dfrac{a+b+2c+1}{1+a}+\dfrac{a+b+2c+1}{1+b}+6\ln(a+b+2c)-2 \\ P=(a+b+2c+1)\left( \dfrac{1}{1+a}+\dfrac{1}{1+b}\right)+6\ln(a+b+2c)-2$$
Ta có: $\dfrac{1}{1+a}+\dfrac{1}{1+b}\ge \dfrac{4}{ab+3}\ge  \dfrac{4}{ab+c(a+b+c)}= \dfrac{4}{(c+a)(c+b)}\ge  \dfrac{16}{(a+b+2c)^2}$
Đặt
$t=a+b+2c=(a+c)+(b+c)\ge 2\sqrt{(a+c)(b+c)}\ge 4 $
Xét hàm....