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

`<=> (4/3 - 1/(a^2 - a + 1)) + (4/3 - 1/(b^2 - b + 1)) + (4/3 -  1/(c^2 - c + 1)) >= 1`

`<=> (4a^2 - 4a + 1)/(3(a^2 - a + 1)) + (4b^2 - 4b + 1)/(3(b^2 - b + 1)) + (4c^2 - 4c + 1)/(3(c^2 - c + 1)) >= 1`

 `<=>  (4a^2 - 4a + 1)/(a^2 - a + 1) + (4b^2 - 4b + 1)/(b^2 - b + 1) + (4c^2 - 4c + 1)/(c^2 - c + 1) >= 3`

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

Ta có :

` (2a - 1)^2/(a^2 - a + 1) + (2b - 1)^2/(b^2 - c + 1) + (2c - 1)^2/(c^2 - c + 1)`

`>= [2(a + b + c) - 3]^2/(a^2 + b^2 + c^2 - (a + b + c) + 3)` (Cauchy Schwaz)

Ta cần chứng minh `:`

`[2(a + b + c) - 3]^2/(a^2 + b^2 + c^2 - (a + b + c) + 3) >= 3`

`<=>[2(a + b + c) - 3]^2 >= 3[a^2 + b^2 + c^2 - (a + b + c) + 3]`

`<=> 4(a + b + c)^2 - 12(a + b + c) + 9 >= 3(a^2 + b^2 + c^2) - 3(a + b + c) + 9`

`<=> 4(a + b + c)^2 - 9(a + b + c) >= 3(a^2 + b^2 + c^2)`

`<=> 4(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) - 3(a^2 + b^2 + c^2) >= 9(a + b + c)`

`<=> (a + b + c)^2 + 6(ab + bc + ca) >= 9(a + b + c)` 

Ta có bđt `:(ab + bc + ca)^2 >= 3abc(a + b + c) = 3(a + b + c)`

`=> (ab + bc + ca) >= sqrt(3(a + b + c))`

`=> (a + b + c)^2 + 6(ab + bc + ca)  >= (a + b + c)^2 + 6sqrt(3(a + b + c))`

Ta sẽ chứng minh `:  (a + b + c)^2 + 6sqrt(3(a + b + c)) >= 9(a + b + c)`

`<=>  (a + b + c)^2 + 6sqrt(3(a + b + c)) - 9(a + b + c) >= 0`

`<=>  (sqrt(a + b + c))^3 + 6sqrt3 - 9sqrt(a + b + c) >= 0` `(1)`

Ta có `:  (sqrt(a + b + c))^3 +  6sqrt3 `

`=  (sqrt(a + b + c))^3 + 3sqrt3 + 3sqrt3 >= 3.root(3)( (sqrt(a + b + c))^3 . 3sqrt3 . 3sqrt3)`

`>= 3root(3)(27)sqrt(a + b + c) = 9sqrt(a + b + c)`

`=> (sqrt(a + b + c))^3 +  6sqrt3 >= 9sqrt(a + b + c)`

Do đó `(1)` đúng

`=>` bđt được chứng minh

Dấu "=" xảy ra khi `:a=b=c=1`