Using AM-GM inequality,consider the following expression :

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)

\(\ge3+2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=3+2+2+2=9\)

Divide both sides of the inequality by a + b + c > 0,we have :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

P/S : a,b,c can't be equal to 0

Use AM-GM for three numbers , we have :

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{\sqrt[3]{abc}}\)

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

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

Use BĐT Cauchy - Schwarz form Engel we have:

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

Equal sign occurs \(\Leftrightarrow a=b=c\)

Using AM-GM inequality,consider the following expression :

(a+b+c)(1a+1b+1c)=1+ab+ac+ba+1+bc+ca+cb+1

=3+(ab+ba)+(bc+cb)+(ac+ca)

≥3+2√ab.ba+2√bc.cb+2√ac.ca=3+2+2+2=9

Divide both sides of the inequality by a + b + c > 0,we have :

1a+1b+1c≥9a+b+c

P/S : a,b,c can't be equal to 0