killpower 發表於 16-5-2009 23:34:11

exercise 6 for ineq.

http://img196.imageshack.us/img196/1733/rw1l.png

bearwing 發表於 17-5-2009 00:02:31

http://img32.imageshack.us/img32/4120/mat5.jpg

killpower 發表於 17-5-2009 00:24:49

本帖最後由 killpower 於 17-5-2009 00:26 編輯

我今次大失預算....[], 出題出得唔好天.

bearwing 發表於 17-5-2009 02:04:21

[]

-終場ソ使者- 發表於 17-5-2009 10:09:16

本帖最後由 -終場ソ使者- 於 17-5-2009 10:10 編輯

我又試下做a先
a)i)
x+y²/x=(√x-y/√x)²+2y
   ≥2y
a)ii)
similarly,
1/x²+1/y²=(1/x-1/y)²+2/xy
    ≥2/xy

bearwing 發表於 23-5-2009 17:57:04

好耐唔見你post
仲有無題目玩[]

-終場ソ使者- 發表於 23-5-2009 20:07:40

a)Prove that x²/(1+x4)≤1/2 for any real number x.
b)If a,b,c>0.Prove that
i)a/b+b/a≥2
ii)(a+b)(1/a+1/b)≥4
iii)a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b)≥6
c)If a,b,c,x,y,z are positive,prove that
√[(x/a+y/b+z/c)(a/x+b/y+c/z)]≥(a+b+c)(x+y+z)/(ax+by+cz)

我係本書搵到呢題,可以做下

bearwing 發表於 23-5-2009 21:29:53

a) x^4 - 2x^2 + 1 >=0
b i) AMGM
b ii) (a+b)(1/a+1/b) =2+ a/b+b/a >= 4
b iii) a(1/b+1/c)+b(1/a+1/c)+c(1/a+1/b) = (a+b+c)(1/a+1/b+1/c) - 3
Consider AMHM (a+b+c)/3 >= 3/(1/a+1/b+1/c)
Hence the result.

c)
>= (sum cyc x)^2 (sum cyc a)^2
Hence the result.

-終場ソ使者- 發表於 23-5-2009 22:17:47

1)
Let a,b,c be positive real numbers.Prove that
abc(a+b+c)≤3/16×[(a+b)(b+c)(c+a)]4/3
2)
Let a,b,c be positive real numbers.Prove that
(a+b+c)/3≤¼{³√[(a+b)²(b+c)²(c+a)²/abc]}
3)
Let a,b,c be positive real numbers.Prove that
³√[(a+b)(b+c)(c+a)/8]≥√[(ab+bc+ca)/3]

都係抄番黎@@

bearwing 發表於 23-5-2009 22:36:52

本帖最後由 bearwing 於 31-5-2009 23:58 編輯

做左

Q2 可唔可以寫返清楚少少
頁: [1] 2
查看完整版本: exercise 6 for ineq.