Lời giải:
Xét hiệu:
\((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-9\)
\(=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1-9\)
\(=\left(\frac{a}{b}+\frac{b}{a}-2\right)+\left(\frac{b}{c}+\frac{c}{b}-2\right)+\left(\frac{c}{a}+\frac{a}{c}-2\right)\)
\(=\frac{(a-b)^2}{ab}+\frac{(b-c)^2}{bc}+\frac{(c-a)^2}{ca}\geq 0, \forall a,b,c>0\)
\(\Rightarrow (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Áp dụng BĐT AM-GM ta có:
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3.\sqrt[3]{abc}.\frac{3}{\sqrt[3]{abc}}=9\)
Dấu " = " xảy ra <=> a=b=c
Akai Haruma: sao thầy không dùng BĐT AM-GM cho nhanh vậy ạ?
