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Simple inequalities.

  1. Oct 25, 2006 #1
    i need to prove the following:
    1)(1+1/n)^n<3 for every n>=3.
    2) (x^n+y^n)/2>=((x+y)/2)^n for every n natural and every x,y>=0.
    3) |a+1/a|>=2 for every a different than 0.

    for the first i thought to use induction and to use the fact of increasing sequence (1+1/n)^n or of the decreasing sequence (1+1/n)^n+1, but i didnt get much far, i also tried to use the fact of 1/n<=1/3, but im stuck.

    for the second i tried by induction: ((x+y)/2)^n+1<=(x^n+y^n)/2(x+y)/2 but im also stuck.
    for the third i tried dissecting it into parts, i.e 0<a<1 a>1 or -1<a<0 a<-1 but i got stuck in a>1.
    i mean for a>1 we can write a=h+1 when h>1 and thus h+1+1/(h+1)>2+1/(h+1)>2 but im having a problem with 0<h<1 then 1<h+1<2 1/2<1/(h+1)<1
    3/2+h<1+h+1/(h+1) but form here i couldnt conclude that it's bigger than 2.
     
  2. jcsd
  3. Oct 25, 2006 #2

    StatusX

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    For the first, start by showing that sequence is bounded above by the other definition of e in terms of the sum of the inverse factorials, and then bound that series by a geometric series.

    For the second, you're on the right track. Try to write what you get from that product as a sum of the expression you want ([itex](x^{n+1}+y^{n+1})/2[/itex]), and some other term, and show this extra term is nonpositive, so that the product is less than or equal to the expression you want.

    Finally, for the last, focus on a>0, and note that your're trying to show there are no solutions to a+1/a<2. This can be rewritten as a quadratic inequality that you can show to have no real solutions.
     
  4. Oct 25, 2006 #3
    but from the prodcut i get [(x^(n+1)+y^(n+1))/2+(x^ny+y^nx)/2]/2 how do i show that it's less than (x^(n+1)+y^(n+1))/2?
    i mean obviously x^ny+xy^n is positive cause x,y>=0
    for the first question, how would i solve this question without the knowledge of e and its definition?

    and about the last question, **** it was that easy. (-:
    thanks.
     
  5. Oct 25, 2006 #4

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    1. Keep going. You know what you want, so just subtract what you found from this, and show what you get is positive (pay attention to factors of 2 and 4).

    2. You don't need to assume anything, just show:

    [tex](1+\frac{1}{n})^n \leq \sum_{m=0}^\infty \frac{1}{m!} [/tex]

    expand the LHS and this shouldn't be too hard.
     
  6. Oct 26, 2006 #5
    ok, thanks.
    btw what's wrong with the latex?
     
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