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
+abc≥21 c3
.
3
答
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...