Question

[Bui Cam Lan Bui]

 Cho a>0; b>0; c>0. Chứng minh bất đẳng thức

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

Answer 1

[Minh Triều]

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

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

Áp dung BĐT cô si cho 2 số không âm ta được:

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)

\(\frac{a}{c}+\frac{c}{a}\ge2\sqrt{\frac{a}{c}.\frac{c}{a}}=2\)

\(\frac{b}{c}+\frac{c}{b}\ge2\sqrt{\frac{b}{c}.\frac{c}{b}}=2\)

Suy ra: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3+2+2+2=9\left(\text{ điều phải chứng minh}\right)\)

Answer 2

[Đinh Tuấn Việt]

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

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

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

Áp dụng tổng hai phân số nghịch đảo lớn hơn hoặc bằng 2 ta có :

\(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge3+2+2+2=9\)

=> ĐPCM