Giải thích các bước giải:
Ta có:
$A=\dfrac{2}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}$
$\to A=\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c}$
$\to A=\dfrac{1^2}{a+b}+\dfrac{1^2}{a+b}+\dfrac{1^2}{a+c}+\dfrac{1^2}{b+c}$
$\to A\ge \dfrac{(1+1+1+1)^2}{(a+b)+(a+b)+(a+c)+(b+c)}$
$\to A\ge \dfrac{16}{3a+3b+2c}$
$\to \dfrac1{3a+3b+2c}\le \dfrac1{16}(\dfrac{2}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{b+c})$
Tương tự
$\dfrac1{3a+2b+3c}\le \dfrac1{16}(\dfrac{2}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{b+c})$
$ \dfrac1{2a+3b+3c}\le \dfrac1{16}(\dfrac{2}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{b+c})$
Cộng vế với vế
$\to VT\le \dfrac1{16}\cdot 4(\dfrac1{a+b}+\dfrac{1}{b+c}+\dfrac1{c+a})$
$\to VT\le \dfrac1{16}\cdot 4\cdot 6$
$\to VT\le \dfrac32$
