对任意x,y＞0,F(x,y)=x+y-a(x+2*(2xy)^0.5)≥0恒成立
即a≤(x+y)/(x+2*(2xy)^0.5)恒成立
而x+y=0.5x+(0.5x+y)≥0.5x+2*(0.5xy)^0.5=0.5x+(2xy)^0.5
故(x+y)/(x+2*(2xy)^0.5)≥(0.5x+(2xy)^0.5)/(x+2*(2xy)^0.5)=0.5,当x=2y时成立
所以a≤0.5