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...