Ta có $\frac{ab}{a+3b+2c}=\frac{ab}{(a+c)+(b+c)+2b}\leq \frac{ab}{9}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{2b})$ ad $\frac{9}{x+y+z}\leq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$TT $\Rightarrow VT \leq \frac{1}{9}(\frac{ab+bc}{a+c}+\frac{ab+ca}{c+b}+\frac{bc+ca}{a+b})+\frac{1}{18}(a+b+c)$
$=\frac{a+b+c}{6}$
Dấu $= \Leftrightarrow a=b=c$