Đặt \(A=x+y+z\)
Theo BĐT cô si, ta có:
\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\frac{1}{xyz}}=9\)
Mà \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{7}{10}\Rightarrow x+y+z\ge9-\frac{7}{10}=\frac{83}{10}\)
Vậy \(A_{min}=\frac{83}{10}\). Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{83}{30}\)