(1)欲证a>(b+c)/2既2a>(b+c)2a=x+y(x+y)>(b+c)(y-c)>(b-x)由题意知在xbcy等比数列中,公比大于1,设为d,得d>1,则d-1>0,d*d>1y=c*d,b=x*d故只须证(c*d-c)>(x*d-x)即c*(d-1)>x*(d-1)c>xc/x>1又c=x*d*dx*d*d/x>1d*d>1,得证.(2)待