Ta có aba+3b+2c≤ab9(1a+b+1b+c+12b)TT ⇒VT≤19(ab+bca+c+ab+cac+b+bc+caa+b)+118(a+b+c) =a+b+c6Dấu =⇔a=b=c
Ta có $\frac{ab}{a+3b+2c}
=\frac{ab}{(a+c)+(b+c)+2b}\leq \frac{ab}{9}(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{2b})
ad\frac{9}{x+y+z}\leq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}TT\Rightarrow VT \leq \frac{1}{9}(\frac{ab+bc}{a+c}+\frac{ab+ca}{c+b}+\frac{bc+ca}{a+b})+\frac{1}{18}(a+b+c)
=\frac{a+b+c}{6}
Dấu= \Leftrightarrow a=b=c$