Áp dụng bđt AM-GM:
\(a^2\sqrt{a}+\frac{1}{\sqrt{a}}\ge2a\)
\(b^2\sqrt{b}+\frac{1}{\sqrt{b}}\ge2b\)
\(c^2\sqrt{c}+\frac{1}{\sqrt{c}}\ge2c\)
Cộng theo vế: \(VT\ge2\left(a+b+c\right)\ge\frac{2}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\) (Cauchy-Schwarz)
\("="\Leftrightarrow a=b=c=1\)