Ta có:
\(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge9\left(a+2b\right)\)
Mặt khác:
\(\left(a+2b\right)^2\le\left(1+2\right)\left(a^2+2b^2\right)\le3\times3c^2\)
\(\Rightarrow\left(a+2b\right)\le3c\)
\(\frac{9}{\left(a+2b\right)}\ge\frac{9}{3c}=\frac{3}{c}\)
\(=VT\ge\frac{3}{c}\left(ĐPCM\right)\)
Dấu "=" xảy ra khi a=b=c=1