`1/a+1/b+1/c>=9/(a+b+c)`
`<=>(ab+bc+ca)/(abc)>=9/(a+b+c)`
`<=>(ab+bc+ca)(a+b+c)>=9abc`
`<=>a^2b+ab^2+abc+abc+b^2c+bc^2+c^2a+abc+ca^2>=9abc`
`<=>a^2b+ab^2+b^2c+bc^2+ca^2+c^2a>=6abc`
Áp dụng bđt cosi ta có:
`a^2b+bc^2>=2abc`
`ab^2+ac^2>=2abc`
`b^2c+a^2c>=2abc`
`=>a^2b+ab^2+b^2c+bc^2+ca^2+c^2a>=6abc`
`=>` ta có đpcm
Dấu "=" `<=>a=b=c`
Áp dụng BĐT Cô-si:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{\sqrt[3]{abc}}\)
Nhân vế với vế:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9\sqrt[3]{abc}}{\sqrt[3]{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) (đpcm)
