设a,b,c为正实数,求证:1a3+1b3+1c3+3abc≥6,并指出等号成立的条件.-数学

题目简介

设a,b,c为正实数,求证:1a3+1b3+1c3+3abc≥6,并指出等号成立的条件.-数学

题目详情

设a,b,c为正实数,求证:
1
a3
+
1
b3
+
1
c3
+3abc≥6
,并指出等号成立的条件.
题型:解答题难度:中档来源:不详

答案

证明:因为a,b,c为正实数,由平均不等式可得class="stub"1
a3
+class="stub"1
b3
+class="stub"1
c3
≥3
3class="stub"1
a3
•class="stub"1
b3
•class="stub"1
c3

即 class="stub"1
a3
+class="stub"1
b3
+class="stub"1
c3
≥class="stub"3
abc
,所以class="stub"1
a3
+class="stub"1
b3
+class="stub"1
c3
+3abc≥class="stub"3
abc
+3abc

而 class="stub"3
abc
+3abc≥2
class="stub"3
abc
•3abc
=6
,所以 class="stub"1
a3
+class="stub"1
b3
+class="stub"1
c3
+3abc≥6

等号成立的条件为
class="stub"1
a
=class="stub"1
b
=class="stub"1
c
class="stub"3
abc
=3abc
,得a=b=c=1.

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