作者Sfly (entangle)
看板Math
标题Re: [分析] 初微(16)
时间Sat Jul 16 12:08:05 2005
※ 引述《PttFund (批踢踢基金)》之铭言：
: Let p, q > 1 with 1/p + 1/q = 1. Show that
: ab ≦ a^p/p + b^q/q
: where a and b are nonnegative.
: Note: 这个就是 Holder inequality, 然後请不要用 Young's inequality
: 来证明.
suppose one of a,b, says a, is not 1.
then b=a^x for some x.
a^(x+1) ≦ a^p/p + a^qx/q
1 ≦ a^(p-1-x)/p + a^((q-1)x-1)/q =: f(x)
then f(x) >0 for x<<0.
now, f'(x)=log a ( -a^(p-1-x)/p + (q-1)a^((q-1)x-1)/q)
f'=0, iff (q-1)a^(qx-p)/q=1/p, ie a^(qx-p)=q/(q-1)p=1
ie. x=p/q.
but f(p/q)=1/p+1/q=1
therefore, f(x) >= 1, for all x,
and the inequality follows.
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