第1个回答 2019-06-16
证明:因为a
+
b
+
c
=
1,把1
=
a
+
b
+
c代入左边可得1/a
+
1/b
+
1/c
=
(a
+
b
+
c)/a
+
(a
+
b
+
c)/b
+
(a
+
b
+
c)/c
=
1
+
b/a
+
c/a
+
a/b
+
1
+
c/b
+
a/c
+
b/c
+
1
=
3
+
(b/a
+
a/b)
+
(c/a
+
a/c)
+
(c/b
+
b/c)
≥
3
+
2√[(b/a)*(a/b)]
+
2√[(c/a)*(a/c)]
+
2√[(c/b)*(b/c)]
=
9,(当且仅当b/a
=
a/b,而且c/a
=
a/c,而且c/b
=
b/c时取等号,即b
=
a
=
c时取等号),所以1/a
+
1/b
+
1/c
≥
9,得证。