Question

已知abc=1,求证:1/[a³（b+c)]+1/[b³（a+c)]+1/[c³（a+b）]大于等于3/2

Answer 1

a、b、c应该都是正数吧。

原式=1/[a³(b+c)]+1/[b³(a+c)]+1/[c³(a+b)]=(abc)/[a³(b+c)]+(abc)/[b³(a+c)]+(abc)/[c³(a+b)]
=1/[a²(1/b+1/c)]+1/[b²(1/a+1/c)]+1/[c²(1/a+1/b)]
根据柯西不等式：
[1/[a²(1/b+1/c)]+1/[b²(1/a+1/c)]+1/[c²(1/a+1/b)]]·[(1/b+1/c)+(1/a+1/c)+(1/a+1/b)]≥(1/a+1/b+1/c)²
即：[1/[a²(1/b+1/c)]+1/[b²(1/a+1/c)]+1/[c²(1/a+1/b)]]·2(1/a+1/b+1/c)≥(1/a+1/b+1/c)²
∴原式=[1/[a²(1/b+1/c)]+1/[b²(1/a+1/c)]+1/[c²(1/a+1/b)]]≥(1/2)(1/a+1/b+1/c)
根据均值不等式：(1/a+1/b+1/c)/3≥[(1/a)·(1/b)·(1/c)]^(1/3)=1，1/a+1/b+1/c≥3
∴原式≥(1/2)×3=3/2