（一）易知，a1＞0,q＞0.原不等式等价于Sn×S(n+2)＜S(n+1)².当q=1时，Sn=na1.显然成立。当q＞0且q≠1时，Sn=a1(q^n-1)/(q-1).由1+q²＞2q可反推出Sn*S(n+2)＜S(n+1)².(二）C=a1/(1-q).a1＞0,0＜q＜1.
