4^x>0,4^y>0
所以4^x+4^y>=2√(4^x*4^y)=2√4^(x+y)
=2√[2^(x+y)]^2
=2*2^(x+y)
=2^(x+y+1)
=2^(x^2+x+1)
x^2+x+1
=(x+1/2)^2+3/4>=3/4
所以2^(x^2+x+1)〉=2^(3/4)
所以log2(4^x+4^y)>=log2 2^(3/4)=3/4
即log2(4^x+4^y)>=3/4