Cassano 發表於 2004-11-4 19:16:50

P.maths inequality 問題

a.)
0 < k < 1
x > 0
prove kx + (1 - k) > x^k

b.)for all a, b > 0 ; prove ka + (1-k)b > (a^k)

我唔識b:icon008::icon008: thxthx~

Saha 發表於 2004-11-4 20:12:09

by partafor0<k<1 and x>0,
kx+(1-k)>x^k
now a,b>0
so ka+(1-k)>a^k
notice that 0<1-k<1
(1-k)b+(1-(1-k))>b^(1-k)
adding
ka+(1-k)b+1>a^k+b^(1-k)
we need to prove
a^k+b^(1-k)-1>a^k*b^(1-k)..(1)
notice that rhs-lhs
(a^k-1)(b^(1-k)-1)<0
so (1) is valid
the result follows

Saha 發表於 2004-11-4 22:19:17

sth wrong

小妖 發表於 2004-11-4 22:32:40

(b)
Let x = a/b

from (a),
k(a/b) + (1-k) > (a/b)^k
ka + (1-k)b > (a/b)^k*b
ka + (1-k)b > a^k*b^(1-k)

yuren 發表於 2004-11-5 13:26:37

死..連part a 都唔識 :icon045::icon116::icon003:

Cassano 發表於 2004-11-5 17:07:23

死..連part a 都唔識 :icon045::icon116::icon003:

D 左佢就ok~

順帶一問: 唔D o既話, 有咩方法做??

beckham312oo 發表於 2004-11-5 18:38:47

呢d 題目...掉返轉黎諗就得....
trivial..
prove:kx + (1 - k) > x^k

即係prove:kx + (1 - k)- x^k >0
Let f(x)=kx + (1 - k)- x^k
見親呢d polynomial...一定係交畀use diffentiation
別無他選
(b)都係trivial...對照返(a) kx + (1 - k) > x^k的結果..
做(b) 一定諗用(a)....
(a)part...(1-k) 後面冇x...
but (b) part .....無啦啦(1-k) 後面有個b 係度...

then 又係掉返轉黎諗..
ka + (1-k)b > (a^k)
=>k(a/b)+(1-k)>(a/b)^k
=>諗用subsitution x=a/b..for all a,b >0<----記得用人d 野前要check condition

remember 上面所有掉返轉黎諗既東西都係寫係草稿紙上....
& pure maths 見親証明題唔識做....try to 掉返轉黎諗...呢d 係我既小小心得

yuren 發表於 2004-11-5 19:03:34

D 左佢就ok~

順帶一問: 唔D o既話, 有咩方法做??
prove
x > x^k - 1    :confused::confused::confused:

yuren 發表於 2004-11-5 19:04:45

呢d 題目...掉返轉黎諗就得....
trivial..
prove:kx + (1 - k) > x^k

即係prove:kx + (1 - k)- x^k >0
Let f(x)=kx + (1 - k)- x^k
見親呢d polynomial...一定係交畀use diffentiation
別無他選
(b)都係trivial...對照返(a) kx + (1 - k) > x^k的結果..
做(b) 一定諗用(a)....
(a)part...(1-k) 後面冇x...
but (b) part .....無啦啦(1-k) 後面有個b 係度...

then 又係掉返轉黎諗..
ka + (1-k)b > (a^k)
=>k(a/b)+(1-k)>(a/b)^k
=>諗用subsitution x=a/b..for all a,b >0<----記得用人d 野前要check condition

remember 上面所有掉返轉黎諗既東西都係寫係草稿紙上....
& pure maths 見親証明題唔識做....try to 掉返轉黎諗...呢d 係我既小小心得
多謝~ :icon065:
你讀緊math? :confused:

Cassano 發表於 2004-11-6 00:41:32

呢d 題目...掉返轉黎諗就得....
trivial..
prove:kx + (1 - k) > x^k

即係prove:kx + (1 - k)- x^k >0
Let f(x)=kx + (1 - k)- x^k
見親呢d polynomial...一定係交畀use diffentiation
別無他選
(b)都係trivial...對照返(a) kx + (1 - k) > x^k的結果..
做(b) 一定諗用(a)....
(a)part...(1-k) 後面冇x...
but (b) part .....無啦啦(1-k) 後面有個b 係度...

then 又係掉返轉黎諗..
ka + (1-k)b > (a^k)
=>k(a/b)+(1-k)>(a/b)^k
=>諗用subsitution x=a/b..for all a,b >0<----記得用人d 野前要check condition

remember 上面所有掉返轉黎諗既東西都係寫係草稿紙上....
& pure maths 見親証明題唔識做....try to 掉返轉黎諗...呢d 係我既小小心得

唔該哂!=3=
其實仲有條唔識, 不過好難打 :icon074::icon074:

舒夫真高 發表於 2004-11-7 13:12:31

其實中間個符號係咪大過等如？

bazookay 發表於 2005-9-9 01:06:13

is there something wrong with part a of this question.....?...coz when x=1, into kx + (1 - k) > x^k.....1>1 which is wrong....is there some missing information in this problem?

Cassano 發表於 2005-9-10 00:09:42

wa嚇死我....成年前架喇wor @@"
我搵唔返喇可能我打漏左, 係 >=

天生射手 發表於 2005-9-10 10:55:12

見x就要用differentiation架啦
x拆兩個unknown a, b就一係put a/b、一係b/a

[ Last edited by 天生射手 on 10-9-2005 at 10:56 AM ]

CYBASTER 發表於 2005-9-10 11:05:24

完全唔識:icon110:
阿sir仲未教到:icon110:
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