Đáp án+Giải thích các bước giải:

Ta có: `a + b + c = 3`

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

Tương tự: `1/(3b + ca) = 1/((b + c)(a + b))`

`1/(3c + ab) = 1/(( a + c)(b + c))`

`-> 1/(3a + bc) + 1/(3b + ca) + 1/(3c + ab) = 1/((a + b)(a + c)) + 1/((b + c)(a + b)) + 1/((a + c)(b + c))`

`= (b + c)/((a + b)(a + c)(b + c)) + (a + c)/((a + b)(a + c)(b + c)) + (a + b)/((a + b)(a + c)(b + c))`

`= (b + c + a + c +  a + b)/((a + b)(a + c)(b + c))`

`= (2 (a + b + c))/((a + b)(a + c)(b + c))`

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

`= 6/((a + b)(a + c)(b + c)) (1)`

Ta lại có:

`6/(\sqrt{ (3a + bc)(3b + ca)(3c + ab)}) = 6/(\sqrt{ (a + b)(a + c)(b + c)(a + b)(a + c)(b + c)})`

`= 6/(\sqrt{ (a + b)^2 . (a + c)^2 . (b + c)^2})`

`= 6/((a + b)(a + c)(b + c)) ( 2)`

Từ `(1)(2) ->` ĐPCM