设a,b,c为正实数,求证:1/a3+1/b3+1/c3+abc≥23.

问题描述:

设a,b,c为正实数,求证:

1
a3
+
1
b3
+
1
c3
+abc≥2
3

证明:因为a,b,c为正实数,由平均不等式可得 1a3+1b3+1c3≥331a3•1b3•1c3,即   1a3+1b3+1c3≥3abc,所以,1a3+1b3+1c3+abc≥3abc+abc,而  3abc+abc≥23abc•abc=23,所以,1a3+1b3+1c3+abc≥23...