\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.\frac{3}{\sqrt[3]{abc}}=9\)
Dấu "=" xảy ra khi \(a=b=c\)
\(S=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}=\frac{4}{10}=\frac{2}{5}\)
\(\Rightarrow S_{min}=\frac{2}{5}\) khi \(x=y=5\)