题干
已知a1，a2，…，an都是正实数，且a1＋a2＋…＋an＝1.求证：
+
+...+
+
≥
.
| a |
| 1 |
| 2 |
| a |
| 1 |
| + |
| a |
| 2 |
| a |
| 2 |
| 2 |
| a |
| 2 |
| + |
| a |
| 3 |
| a |
| n |
| - |
| 1 |
| 2 |
| a |
| n |
| - |
| 1 |
| + |
| a |
| n |
| a |
| n |
| 2 |
| a |
| n |
| + |
| a |
| 1 |
| 1 |
| 2 |
答案(点此获取答案解析）
证明：根据柯西不等式，得左功＝(a1+a2)+(a2+a3)+...+(an-1+an)+(an+a1)
