Question

a. Cho 3 số dương a, b, c có tổng bằng 1. Chứng minh rằng: 

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

Answer 1

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

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

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

Vậy \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\ge3+2+2+2=9\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}a=b=c\\a+b+c=1\end{cases}}\Rightarrow a=b=c=\frac{1}{3}\)

Answer 2

Áp dụng BĐT AM-GM (Cô si) cho hai số dương,ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}=9^{\left(đpcm\right)}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\\a+b+c=1\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b=c\\a+b+c=1\end{cases}}\Leftrightarrow a=b=c=\frac{1}{3}\)