Question

Cho a, b, c > 0 và a+b+c = 3. Tìm Min \(S=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Answer 1

Do a ; b ; c > 0 , áp dụng BĐT Cô - si cho 3 số , ta có :

\(a+b+c\ge3\sqrt[3]{abc};\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

\(\Rightarrow3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)