{（an）*（an+1）}是公比为q(q>0)的等比数列
所以[（an+1）*（an+2）]/[（an）*（an+1）]=q
an+2/an=q
an+2=q*an
an+3=q*an+1
(an)*(an+1)+(an+1)*(an+2)=(an)*(an+1)+(an+1)*(an*q)=(1+q)*(an)*(an+1)
(an+2)*(an+3)=(an)*q*(an+1)*q=q^2*(an)*(an+1)
所以(1+q)*(an)*(an+1)>q^2*(an)*(an+1)
q^2-q-10
所以0