设甲到乙长S; 则:t1=S/a,t2=S/b,t1>t2v=2S/(t1+t2)=2S/[S/a+S/b]=2/(1/a+1/b)就是a与b的调和平均数;=2ab/(a+b)因为:a+b>2√(ab)>0;所以:1/(a+b)<1/2√(ab); 2ab/(a+b)<2ab/2√ab=√abv=2ab/(a+b)<√ab<(a+b)/2显然v>a;所以选A
v=2S/[S/a+S/b]=2ab/(a+b)假设a=1,b=2,则v=4/3,比较可得选A