Chứng minh rằng: $\frac{1}{3}$ ≤ $\frac{x^{2} +x+1}{x^{2}-x+1}$ ≤3

Đặt `A=(x^2+x+1)/(x^2-x+1)`

Vậy ta cần chứng minh `1/3 \le A\le 3`

Xét `A-1/3`

`= (x^2+x+1)/(x^2-x+1)-1/3`

`= (3x^2+3x+3 - x^2+x-1)/(3(x^2-x+1))`

`= (2x^2 +4x+2)/(3(x^2-2.x. 1/2+1/4+3/4))`

`= (2(x+1)^2)/(3[(x-1/2)^2+3/4])>=0∀x`

`->A-1/3>=0∀x`

`->1/3 \le A(1)`

Xét `3-A`

`=3-(x^2+x+1)/(x^2-x+1)`

`= (3x^2-3x+3 - x^2-x-1)/(x^2-x+1)`

`=(2x^2 - 4x+2)/(x^2-2.x. 1/2+(1/2)^2+3/4)`

`= (2(x-1)^2)/((x-1/2)^2+3/4)>=0∀x`

`->3-A>=0∀x`

`->A\le 3(2)`

`(1),(2)->1/3\le A\le 3`

$\to$đpcm