cho các số thực a;b;c;d thỏa a+b+c+d=2. Chứng minh $\frac{a}{a^2-a+1}+$ $\frac{b}{b^2-b+1}+$ $\frac{c}{c^2-c+1}+$ $\frac{d}{d^1-d+1}$ =<$\frac{8}{3}$

Đặt `{(x=a-1/2),(y=b-1/2),(z=c-1/2),(t=d-1/2):}=>x+y+z+t=0`

`BĐT<=>\frac{2(2x+1)}{4x^2+3}+\frac{2(2y+1)}{4y^2+3}+\frac{2(2z+1)}{4z^2+3}+\frac{2(2t+1)}{4t^2+3}<=8/3`

`<=>\frac{(2x-1)^2}{4x^2+3}+\frac{(2y-1)^2}{4y^2+3}+\frac{(2z-1)^2}{4z^2+3}+\frac{(2t-1)^2}{4t^2+3}>=4/3` (1)

Ta có:

`4x^2+3=3x^3+3+(y+z+t)^2<=3x^2+3+3(y^2+z^2+t^2)=3(x^2+y^2+z^2+t^2+1)`

`=>\frac{(2x-1)^2}{4x^2+3}>=\frac{(2x-1)^2}{3(x^2+y^2+z^2+t^2+1)}`

Tương tự cho hai BĐT còn lại rồi cộng theo vế

`VT_{(1)}>=\frac{(2x-1)^2+(2y-1)^2+(2z-1)^2+(2t-1)^2}{3(x^2+y^2+z^2+t^2+1)}`

`=\frac{4(x^2+y^2+z^2+t^2+1)-4(x+y+z+t)}{3(x^2+y^2+z^2+t^2+1)}`

`=\frac{4(x^2+y^2+z^2+t^2+1)}{3(x^2+y^2+z^2+t^2+1)}=4/3=VP_{(1)}`