Bài làm:
Ta có: \(a+b^2+c^3=\left(a+\frac{1}{a}\right)+\left(b^2+\frac{1}{b}+\frac{1}{b}\right)+\left(c^3+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)-\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right)\)
\(\ge2.1+3.1+4.1-6=3\)
Dấu "=" <=> \(\hept{\begin{cases}a^2=1\\b^3=1\\c^4=1\end{cases}\Rightarrow a=b=c=1}\)
Học tốt!!!!