Pure maths Inequality~~~

(C)
(1)先證明 a^n+b^n>= 2^(n+1)
n=1 即Q(b) holds
a^n+b^n >= 2√(a^n b^n)= 2(ab)^(n/2) (note: 1/a+1/b=1=> a+b=ab)
= 2(a+b)^(n/2) >= 2* 4^(n/2) = 2^(n+1)
by M.I. => a^n + b^n >= 2^(n+1) ------(A)
(2)證明本題
(a+b)^(n+1) - a^(n+1) - b^(n+1)
= (a+b)(a+b)^n - a^(n+1)- b^(n+1)
>= (a+b)[ a^n+b^n + 2^(2n) - 2^(n+1) ] - a^(n+1) - b^(n+1)
= a^(n+1)+b^(n+1) + ab^n+ba^n +(a+b)[2^(2n)-2^(n+1)] - a^(n+1)-b^(n+1)
= ab[ a^(n-1)+b^(n-1)] + 4[ 2^(2n)- 2^(n+1)] (note: ab= a+b>= 4)
>= 4* 2^n + 4[2^(2n)- 2^(n+1)] ----( by (A) )
= 2^(2n+2) - 2^(n+2)
by MI. QED.

HI! 祝考試順利,好成績要告知喔!


2009-03-29 20:38:21 補充：
note: 樓上的作法有點問題
In fact, (a+b)^n - a^n - b^n 有可能等於 2^(2n)- 2^(n+1) ( when a=b=2 )

2009-03-29 20:40:57 補充：
a^n+b^n >= 2^(n+1) 沒有用 MI證明! so, (A)式不用寫 " by M.I. "

2009-03-29 21:06:43 補充：
倒數第3式
= ab[ a^(n-1)+b^(n-1)] + 4[ 2^(2n)- 2^(n+1)]
更正為:
>= ab[ a^(n-1)+b^(n-1)] + 4[ 2^(2n)- 2^(n+1)]

2009-03-30 19:54:44 補充：
如何從part A中得知a^n + b^n >= 2^(n + 1) ?
1/a+1/b=1 =>ab=a+b >= 4
a^n+b^n >= 2 √(ab)^n >= 2* √4^n= 2^(n+1)
so, a^n+b^n >= 2^(n+1) (for all n>= 1)

2009-03-30 21:19:51 補充：
MI之最後兩步:
ab[a^(n-1)+b^(n-1)] (Note: ab=a+b >= 4, a^(n-1)+b^(n-1)>= 2^n )
>= 4*2^n

c)

when n=1 ,

(a+b)^1-a^1-b^1=0

so it is true for n =1 ,

Assume it is true for n= k , where k belongs to positive integer ,

i.e. (a+b)^k-a^k-b^k >= 2^(2k)-2^(k+1)

Cosinder when n=k+1

(a+b)^(k+1) -a^(k+1) -b^(k+1)

>= (a+b)[2^(2k)-2^(k+1)+a^k+b^k]-a^(k+1) -b^(k+1)
>=4[2^(2k)-2^(k+1)-a^k-b^k]+a^(k+1) +b^(k+1)
=2^(2k+2)-2^(k+2)-2^(k+2)+(a-4) a^k +(b-4)b^k
>=2^(2k+2)-2^(k+2) -2^(k+2) -(ab)[a^(k-1)+b^(k-1)]
>=2^(2k+2)-2^(k+2)

So it is true for n=k+1

By Mathematical Induction , It is true for all natural number n

since a^2 + b^2 >= 2ab
( a^2 + b^2 )/ab >= 2 for ab>0
a/b + b/a >= 2



a+b = (a+b)*(1/a + 1/b)
a+b = 1+ 1 + b/a + a/b
a+b = 2 + b/a + a/b >= 2+ 2 =4

a) (b/a + a/b)/2 >= √(b/a)(a/b)


b) From 1/a + 1/b = 1,

we have : a+b = ab

By a^2 + b^2 >= 2ab

we have : (a+b)^2 - 2ab >= 2ab

(ab)^2 - 2ab >=2ab

ab > = 4

So a+b >= 4

c) Cannot see the All Question.