Question

CMR nếu a+b+c=1 và a.b.c>0 thì ( \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \)) >= 9

Answer 1

Lớp 9 chưa học cauchy thì làm cách này nha :v

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

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

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

\(-->đpcm\) \("="\) khi \(a=b=c=\dfrac{1}{3}\)

Answer 2

áp dụng cauchy-schwarz dạng engel ta có :

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