Question

局部柯西不等式求解释

Answer 1

用1减不等式两边.左边得(1+xy)/((1+x)(1+y)),右边得(1+xy)/(1+√(xy))².

于是不等式化为(1+√(xy))² ≤ (1+x)(1+y),这就是典型的Cauchy不等式了.

证明顺过来写就是这样:

∵x,y > 0,由Cauchy不等式有(1+√(xy))² ≤ (1+x)(1+y).

∴(1+xy)/((1+x)(1+y)) ≤ (1+xy)/(1+√(xy))².

∴(x+y)/((1+x)(1+y)) = 1-(1+xy)/((1+x)(1+y)) ≥ 1-(1+xy)/(1+√(xy))² = 2√(xy)/(1+√(xy))².