Question

Cho a,b,c > 0 và a + b + c = 1. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)

Answer 1

Áp dụng AM-GM

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

\(\rightarrow1.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

vậy ta có điều phải chứng minh

Dấu "=" \(a=b=c=\dfrac{1}{3}\)

Answer 2

Áp dụng svac-xơ:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=9\)

Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{3}\)

C2: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)

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

\(\ge3+2+2+2=9\) (theo cosi)

Dấu = xảy ra <=>\(a=b=c=\dfrac{1}{3}\)