ta có :
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{b+c}$\geq$\frac{4^{2}}{2a+3b+3c}$
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+c}$\geq$\frac{4^{2}}{2b+3a+3c}$
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}$\geq$\frac{4^{2}}{2c+3a+3b}$
Cộng vế theo vế ta có :
$4(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})$\geq$4^{2}(\frac{1}{2a+3b+3c}+\frac{1}{2b+3a+3c}+\frac{1}{2c+3a+3c})$
~~> $\frac{1}{2}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})$\geq$\frac{1}{4}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})$\geq $VT_{(1)}$ (đpcm)
P/s : đề bài không chặt chẽ nhé