Question

Tìm GTNN của \(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

biết \(a;b;c>0\)

Answer 1

Ta có:

\(A=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\frac{9}{a+b+c}=9\)

Dấu = xảy ra khi a = b = c

Answer 2

ap dung nếu cần c/m:\(t+\frac{1}{t}\ge2\) mọi t>0 đẳng thức khi t=1

\(\ge9\) khi a=b=c

Answer 3

Áp dung BĐT Cô - si

\(_{\Rightarrow\hept{\begin{cases}a+b+c\ge\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{cases}}}3\sqrt[3]{abc}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc.\frac{1}{abc}}\)

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

Vậy GTNN của  \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=9\)

Answer 4

Đáp án : = 9