\(a+2b=1.a+\sqrt{2}.\sqrt{2}b\le\sqrt{\left(1+2\right)\left(a^2+2b^2\right)}\le\sqrt{3.3c^2}=3c\)
\(\Rightarrow a+2b\le3c\)
\(\Rightarrow\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\) (đpcm)
Dấu "=" khi \(a=b=c\)