设a,b是方程4x^2-4mx+(m+2)=0的两个实根(m∈R),则a^2+b^2的最小值

有实根所以
(-4m)^2-16(m+2)>=0
m^2-m-2>=0
(m-2)(m+1)>=0
m>=2,m<=-1
a+b=-(-4m)/4=m
ab=(m+2)/4
a^2b^2=(a+b)^2-2ab=m^2-(m+2)/2
=m^2-m/2-1
=(m-1/4)^2-17/16
m>=2,m<=-1
所以m=-1时,最小值=1/2