Results 1 to 5 of 5

Math Help - Proving (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.

  1. #1
    Newbie
    Joined
    Jan 2011
    Posts
    11

    Proving (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.

    Let x1,x2,...,xn be positive numbers satisfying
    (x1)(x2)(x3)...(xn)=1.
    Show that for any t>=0,
    (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.
    Last edited by mr fantastic; March 7th 2011 at 10:33 AM. Reason: Re-titled.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32
    We can show that \dfrac{\left(1+\frac t{x_i}\right)(1+tx_i)}{(1+t)^2}\geq 1 hence we have either (1+tx_1)\ldots(1+tx_n)\geq (1+t)^n or \left(1+\dfrac t{x_1}\right)\ldots \left(1+\dfrac t{x_n}\right) \geq (1+t)^n. If it is the second case, put for t\neq 0 f(t) =\left(1+\dfrac t{x_1}\right)\ldots \left(1+\dfrac t{x_n}\right)-(1+t)^n and compute f\left(\dfrac 1t\right).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,517
    Thanks
    771
    Nice. Could you explain one thing? If I understand correctly, you claim that \dfrac{\prod_{i=1}^n(1+tx_i)}{(1+t)^n}\cdot\dfrac{  \prod_{i=1}^n(1+t/x_i)}{(1+t)^n}\ge1, from where either the first or the second factor is \ge 1. Theoretically, the choice of which factor exceeds 1 can depend on t. However, it seems that later you require that either
    \prod_{i=1}^n(1+tx_i)\geq (1+t)^n for all t or \prod_{i=1}^n(1+t/x_i)\geq (1+t)^n for all t. This allows you substituting 1/t for t in case the second alternative is true. But is it possible, say, that the second alternative is true for t = 5, but the first alternative is true for t = 1/5?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    675
    Thanks
    32
    Quote Originally Posted by emakarov View Post
    However, it seems that later you require that either
    \prod_{i=1}^n(1+tx_i)\geq (1+t)^n for all t or \prod_{i=1}^n(1+t/x_i)\geq (1+t)^n for all t. This allows you substituting 1/t for t in case the second alternative is true. But is it possible, say, that the second alternative is true for t = 5, but the first alternative is true for t = 1/5?
    You're right, what I did cannot be correct because I didn't use the fact that x_1\ldots x_n=1.
    I think we can show the result by induction. It's not difficult for n=2 and it's true for n and x_1\ldots x_{n+1}=1 were x_i is positive.
    We assume x_{n+1}\leq 1 and x_n\geq 1 (by symmetry ). Let y_i=x_i if i\leq n-1 and y_n=x_nx_{n+1}. We have \prod_{i=1}^n(1+y_it)\geq (1+t)^n for any t\geq 0 hence \prod_{i=1}^{n-1}(1+x_it)(1+x_nx_{n+1}t)\geq (1+t)^n . Now we have to show that
    (1+x_nx_{n+1}t)(1+t)\leq (1+x_{n+1}t)(1+x_nt). If we compute the difference we get (1+x_{n+1}t)(1+x_nt) -(1+x_nx_{n+1}t)(1+t)=t(x_{n+1}+x_n-x_nx_{n+1}-1)t(1-x_n)(x_{n+1}-1) which is positive.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Mar 2008
    Posts
    934
    Thanks
    33
    Awards
    1
    Quote Originally Posted by chris86 View Post
    Let x1,x2,...,xn be positive numbers satisfying
    (x1)(x2)(x3)...(xn)=1.
    Show that for any t>=0,
    (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.
    Let
    f(x) = \ln(1 + e^x).
    It's easy to show that the second derivative is positive, so f is convex. By Jenson's Inequality,

    f((a_1 + a_2 + \dots + a_n)/n) \leq f(a_1)/n + f(a_2)/n + \dots + f(a_n)/n

    I.e.,
    \ln(1 + e^{(a_1 + a_2 + \dots + a_n)/n}) \leq \ln(1 + e^{a_1})/n + \ln(1 + e^{a_2})/n + \dots + \ln(1 + e^{a_n})/n
    for any a_1, a_2, \dots , a_n.

    Exponentiating,
    1 + e^{(a_1 + a_2 + \dots + a_n)/n} \leq (1+ e^{a_1})^{1/n} (1+ e^{a_2})^{1/n} \cdots (1+ e^{a_n})^{1/n} .

    Now let a_i = \ln(x_i t), so
    e^{(a_1 + a_2 + \dots + a_n)/n} = (x_1 x_2 \cdots x_n)^{1/n} t = t
    and the inequality becomes
    1+t \leq (1 + x_1 t)^{1/n} (1 + x_2 t)^{1/n} \cdots (1 + x_n t)^{1/n}.

    Finally, raise both sides to the nth power.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving y^2=-4x
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 27th 2010, 07:48 AM
  2. Proving CRT
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: April 13th 2010, 10:14 PM
  3. proving
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: April 11th 2010, 12:49 AM
  4. Please me proving this.
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 27th 2010, 11:29 PM
  5. Proving an identity that's proving to be complex
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: July 21st 2009, 01:30 PM

Search Tags


/mathhelpforum @mathhelpforum