3/2*5/4*7/6*9/8*...2n+1/2n>根号下n+1?怎么证明?

证明：当n=1时
3/2>√2成立
假设当n=k时命题成立 即 3/2*5/4*7/6*9/8*...2k+1/2k>√(k+1)成立
即（3/2*5/4*7/6*9/8*...2k+1/2k）^2>k+1成立
即要证当 n=k+1时
3/2*5/4*7/6*9/8*...2k+1/2k*(2k+3)/(2k+2)>√(k+2)
两边平方
得到[3/2*5/4*7/6*9/8*...2k+1/2k*(2k+3)/(2k+2)]^2>(k+2)
因为（3/2*5/4*7/6*9/8*...2k+1/2k）^2>k+1
（3/2*5/4*7/6*9/8*...2k+1/2k）^2*[(2k+3)/(2k+2)]^2>(k+1)
[(1+1/2(k+1)]^2=(k+1)[1+1/(k+1)+1/4(k+1)^2]=k+2+1/4(k+1)^2>k+2
n=k+1时命题成立
得证