Ta có:

`a+b+c+ab+bc+ca=6abc`

`=> 1/(bc) + 1/(ac) + 1/(ab) + 1/c + 1/a + 1/b = 6`

Áp dụng bất đẳng thức Cô-si cho hai số dương:

`{(1/a^2 + 1/b^2 >= 2\sqrt{1/a^2 . 1/b^2} = 2\sqrt{ 1/(a^2 b^2)} = 2 . 1/(ab)), (1/b^2 + 1/c^2 >= 2\sqrt{1/b^2 . 1/c^2} = 2\sqrt{ 1/(b^2 c^2)} = 2 . 1/(bc)), (1/c^2 + 1/a^2 >= 2\sqrt{1/c^2 . 1/a^2} = 2\sqrt{ 1/(c^2 a^2)} = 2 . 1/(ca)):}`

`=> 2 . (1/a^2 + 1/b^2 + 1/c^2) >= 2 . (1/(ab) + 1/(bc) + 1/(ca))`

`<=> 1/a^2 + 1/b^2 + 1/c^2 >= 1/(ab) + 1/(bc) + 1/(ca) (1)`

`{(1/a^2 + 1 >= 2\sqrt{1/a^2 +1} = 2 . 1/a),(1/b^2 + 1 >= 2\sqrt{1/b^2 +1} = 2 . 1/b),(1/c^2 + 1 >= 2\sqrt{1/c^2 +1} = 2 . 1/c):}`

`=> 3+ 1/a^2 + 1/b^2 + 1/c^2  >= 2 . (1/a + 1/b + 1/c)`

`<=> 1/2 . (1/a^2 + 1/b^2 + 1/c^2) + 3/2 >= 1/a + 1/b + 1/c (2)`

`(1)(2)=> 3/2 .(1/a^2 + 1/b^2 + 1/c^2) + 3/2 >= 1/(ab) + 1/(bc) + 1/(ac) + 1/a + 1/b + 1/c`

`<=> 3/2 .(1/a^2 + 1/b^2 + 1/c^2) + 3/2 >= 6`

`<=> 1/a^2 + 1/b^2 + 1/c^2 >= 3`

Dấu "=" xảy ra `<=>{(1/a^2 = 1/b^2 = 1/c^2 ),(a+b+c+ab+bc+ca=6abc):} <=>a=b=c`

`=>đpcm`.