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Jensen's inequality and a question concerning e.

  1. Oct 2, 2006 #1
    1) find the least value a such that (1+1/x)^(x+a)>e
    for all positive x.
    2) let a,b be two positive numbers, p and q any nonzero numbers p<q.
    prove that [ta^p+(1-t)b^p]^1/p<=[ta^q+(1-t)b^q]^1/q.(0<t<1)

    for the first question im given the hint that [1+1/x]^x+1 decreases monotonically and [1+1/x]^x increases monotonically to the limit e at infinity, so basically we have here a function of a that as a increases the function decreases monotonically, so i need to take the derivative of the rhs and find for a between 0 and 1, which is the least value of a.
    the problem is that derivative that i got wrt a is:
    (1+1/x)^(x+a)*ln(1+1/x) and i need to equate this to 0 because x>0 ln(1+1/x)>0 and thus this derivative is greater than zero for all a.
    so im stuck here.

    for the second question i thought to take the derivative of the function [ta^n+(1-t)b^n]^s when s is the variable and aftwerwards when the variable is n, i got that those two functions increase monotonically, i tried to put into n and s p and q, and i got 4 inequalities, i thought from these four i could conclude the required inequality but didnt succeed.
     
  2. jcsd
  3. Oct 2, 2006 #2

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    You can tell if the function approaches e from above or below by seeing if the derivative approaches 0 from above or below. If it approaches zero from above, the function must be less than e for sufficiently large x, so you can exclude the corresponding values of a. If it's from below, you still need to verify that the function never crosses e and then comes back up.
     
  4. Oct 3, 2006 #3
    what about jensen's inequality?
     
  5. Oct 3, 2006 #4

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    Use the convexity of x^(q/p).
     
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