→ erre : so? 05/24 18:48
→ Desperato : 我證錯了XD 看隔壁 05/24 20:26
給定 0<a,a',b,b'<1, a+a'<=1, b+b'<=1
b<a, 1/2<a', 0<m<n, m,n為自然數
(1) b^n*b'^m < a^m*a'^n
(2) b*b' <= a*a'
(Thm1) (1) iff (2)
(pf) Note that b/a' < 1
"=>" if b/a' > a/b', then b/a' => (a/b')^r
for some r=m/n close to 1
thus (b/a')^n => (a/b')^m
"<=" (b/a')^n < (b/a')^m <= (a/b')^m
(Thm2) (2) does not hold
(pf) a=0.25, a'=0.6, b=0.2, b'=0.8
is a counterexample
Therefore (1) fails.
one can use Thm1 to obtain
a counterexample for (1)
p.s. if a+a'=b+b', then (2) holds
p.s. if no 1/2<a', then b/a' may => 1
which make (1) fails when m, n is big
----
Sent from BePTT
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 180.204.65.139
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1558633606.A.94B.html