P.maths inequality 問題

Cassano

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) [b^(1-k)]

我唔識b     thxthx~

Saha

by parta  for0<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

sth wrong

小妖

(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

死..連part a 都唔識     

Cassano

[QUOTE=yuren]死..連part a 都唔識     [/QUOTE]

D 左佢就ok~

順帶一問: 唔D o既話, 有咩方法做??

beckham312oo

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) [b^(1-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

[QUOTE=Cassano]D 左佢就ok~

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

yuren

[QUOTE=beckham312oo]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) [b^(1-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[/QUOTE]
多謝~
你讀緊math?

Cassano

[QUOTE=beckham312oo]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) [b^(1-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[/QUOTE]

唔該哂!=3=
其實仲有條唔識, 不過好難打   

舒夫真高

其實中間個符號係咪大過等如？

bazookay

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

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

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

完全唔識
阿sir仲未教到