`T=(4a-b)/(1+b)-(2a-2b+1)/(1+2a)-(6a+3b-6077)/(6a+3b)`

`=(4a)/(b+1)+(2b)/(2a+1)+6077/(6a+3b)-3`

`=(8a^2+4a+2b^2+2b)/((2a+1)(b+1))+6077/(6a+3b)-3`

`=(4a+2b+1)/((2a+1)(b+1))+6077/(6a+3b)-3`

`>=(8(2a+b)+4)/(2a+b+2)^2+6077/(3*(2a+b))-3`

Đặt: `2a+b=x`

Áp dụng bất đẳng thức Bunhiacopxki:

`2a+b=\sqrt{(1*2a+1*b)^2}<=\sqrt{(1^2+1^2)(4a^2+b^2)}=\sqrt{8a^2+2b^2}=1`

Do đó: `0<x<=1`

`T=(8x+4)/(x+2)^2+6077/(3x)-3`

`=((1-x)(6081x^2+24304x+24308))/(3x(x+2)^2)+2024`

`>=2024`

Dấu "`=`" xảy ra khi: `a=1/4,b=1/2`