Question

Cho 3 số a , b , c có tổng bằng 1 . Chứng minh rằng : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Answer 1

Từ : \(a+b+c=1\) \(\Rightarrow\hept{\begin{cases}\frac{1}{a}=1+\frac{b}{a}+\frac{c}{a}\\\frac{1}{b}=1+\frac{a}{b}+\frac{c}{b}\\\frac{1}{c}=1+\frac{a}{c}+\frac{b}{c}\end{cases}}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

\(\ge3+2+2+2=9\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Answer 2

Bổ sung a,b,c dương vào đê

Cách 1:

Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

Ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{4}{a+b}+\frac{1}{c}\ge\frac{9}{a+b+c}=9\)

Dấu "=" xảy ra tại a=b=c=¹/3 

Cách 2:

Áp dụng BĐT Cô si ta có:

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

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

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\frac{1}{abc}}=9\)

Mà \(a+b+c=1\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\)

Cách 3:

Xét:\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)

\(\ge3+2\sqrt{\frac{a}{b}\cdot\frac{b}{a}}+2\sqrt{\frac{b}{c}\cdot\frac{c}{b}}+2\sqrt{\frac{c}{a}\cdot\frac{a}{c}}\)

\(=3+2+2+2\)

\(=9\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\) vì a+b+c=1