Results 1 to 3 of 3

Math Help - Inequalities

  1. #1
    Member
    Joined
    Jul 2011
    Posts
    140

    Inequalities

    Can anyone help me confirm if I have solved this correctly?

    Many thanks.

    Q.
    (1 + 2x)n > 1 + 2nx, for x > 0 & n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    (1 + 2x)1 = 1 + 2x '&' 1 + 2(1)x = 1 + 2x
    Since 1 + 2x > 1 + 2x, n is true.

    Step 2:
    Assume the statement is true for n = k, i.e. assume (1 + 2x)k > 1 + 2kx,
    We must now show that the statement is true for n = k + 1,
    i.e. (1 + 2x)k+1 > 1 + 2(k + 1)x = 1 + 2kx + 2x
    (1 + 2x)k+1 = (1 + 2x)1(1 + 2x)k > (1 + 2x)(1 + 2kx)...((1 + 2x)k > 1 + 2kx...assumed)
    If (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x then (1 + 2x)k+1 > 1 + 2kx + 2x
    if (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x
    if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x) > 0
    if 2kx2 + 2kx + 2x + 1 - 1 - 2kx - 2x > 0
    if 2kx2 > 0
    if x2[2k] > 0...true when x > 0 & n > 1
    (1 + 2x)k+1 > 1 + 2kx + 2x
    Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n > 1, n \in \mathbb{N}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,548
    Thanks
    1418

    Re: Inequalities

    Quote Originally Posted by GrigOrig99 View Post
    Can anyone help me confirm if I have solved this correctly?

    Many thanks.

    Q.
    (1 + 2x)n > 1 + 2nx, for x > 0 & n \in \mathbb{N}

    Attempt: Step 1: For n = 1...
    (1 + 2x)1 = 1 + 2x '&' 1 + 2(1)x = 1 + 2x
    Since 1 + 2x > 1 + 2x, n is true.

    Step 2:
    Assume the statement is true for n = k, i.e. assume (1 + 2x)k > 1 + 2kx,
    We must now show that the statement is true for n = k + 1,
    i.e. (1 + 2x)k+1 > 1 + 2(k + 1)x = 1 + 2kx + 2x
    (1 + 2x)k+1 = (1 + 2x)1(1 + 2x)k > (1 + 2x)(1 + 2kx)...((1 + 2x)k > 1 + 2kx...assumed)
    If (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x then (1 + 2x)k+1 > 1 + 2kx + 2x
    if (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x
    if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x) > 0
    if 2kx2 + 2kx + 2x + 1 - 1 - 2kx - 2x > 0
    if 2kx2 > 0
    if x2[2k] > 0...true when x > 0 & n > 1
    (1 + 2x)k+1 > 1 + 2kx + 2x
    Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n > 1, n \in \mathbb{N}
    I'd do it like this...

    \displaystyle \begin{align*} (1 + 2x)^{k + 1} &= (1 + 2x)(1 + 2x)^k \\ &\geq (1 + 2x)(1 + 2kx) \\ &= 1 + 2kx + 2x + 2kx^2 \\ &\geq 1 + 2kx + 2x \end{align*}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2011
    Posts
    140

    Re: Inequalities

    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 12th 2009, 07:33 PM
  2. Inequalities Q
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: September 4th 2009, 02:10 AM
  3. Help me! inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 31st 2008, 07:56 PM
  4. pdf about inequalities
    Posted in the Math Forum
    Replies: 0
    Last Post: August 26th 2008, 04:09 AM
  5. Inequalities help...
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 13th 2008, 11:41 AM

Search Tags


/mathhelpforum @mathhelpforum