Re: [微積] 證明不等式
※ 引述《zako1113 (那個人)》之銘言：
: 被學生問這道題, 但是完全沒有頭緒
: 背景是只有1科微積分
: f(x) defined on [0,1], f(x)>0, f"(x)<=0,
: given a constant 0<M<0.5, show that
: for any t such that M < t < 1-M, any s belong to (0,1),
: f(t) >= M*f(s).
Since f"(x)<=0, f is a concave function.
i.e. f( (1-w)*x+w*y ) >= (1-w)*f(x)+w*f(y).
for any w belong to (0,1) and 0 <= x,y <= 1.
Now fixed M, t and s satisfying above conditions,
let w = M, y = s and x = (t-M*y)/(1-M).
It is easy to show that 0 <= x <= 1.
Hence f(t)=f( (1-M)*x+M*y ) >= (1-M)*f(x)+M*f(y) >= M*f(y) = M*f(s).
The last inequality is from f(x)>0.
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.104.91
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1450261534.A.45D.html
推
12/16 19:34, , 1F
12/16 19:34, 1F
討論串 (同標題文章)