`#AkaShi`
`a) a^2+b^2+4 >= ab+2b+2a`
`⇔2(a^2+b^2+4) >= 2(ab+2b+2a)`
`⇔2a²+2b²+8 >= 4ab+4b+4b`
`⇔2a²+2b²+8-4ab-4a-2b >= 0`
`⇔a²+a²+b²+b²+4+4-4ab-4b-2b >=0`
`⇔(a²-2ab+b²)+(a²-4a+4)+(b²-4b+4) >= 0`
`⇔[(a)^2-2.a.b+(b)^2]+[(a)^2-2.a.2+(2)^2]+[(b)^2-2.2.b+(2)^2] >= 0`
`⇔(a-b)^2+(a-2)^2+(b-2)^2 >= 0` (Luôn đúng)
Vậy `a^2+b^2+4 >= ab+2b+2a`