Results 1 to 7 of 7

Math Help - Proving a formula

  1. #1
    Member mohammadfawaz's Avatar
    Joined
    Feb 2010
    From
    Lebanon - Beirut
    Posts
    100

    Proving a formula

    Hello all,

    I need help proving the following:
    if a,b,c,... are n real positive number such that a+b+c+... = 1, then
    a^2 + b^2 + c^2 + ... >=1/n

    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    Hello,
    denote all these numbers a_1, a_2, ... a_n, with the property :

    \sum_{k = 1}^{n} a_k = 1

    Now note that when a_k = \frac{1}{n}, we have a_k^2 = \frac{1}{n^2}, and it follows that :

    \sum_{k = 1}^{n} a_k^2 = \sum_{k = 1}^{n} \frac{1}{n^2} = n \times \frac{1}{n^2} = \frac{1}{n}

    This is not a proof in itself : it is there to help you grasp what this problem means. All you have to prove is that there is no solution for a_k where \sum_{k = 1}^{n} a_k^2 < \frac{1}{n}.

    Here is an idea : say a_1 = \frac{1}{x}. Therefore, the following holds : 0 < a_2 < 1 - \frac{1}{x}, and so on following the same logic with a_3, a_4, .... Take the squares of these inequalities, and see where you get You will very likely stumble upon a difficulty, that you will probably be able to express in mathematical terms and that will lead to the solution to your problem
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mohammadfawaz's Avatar
    Joined
    Feb 2010
    From
    Lebanon - Beirut
    Posts
    100
    Hey..
    Thank you for the reply...
    I did what we said: I bounded each number between 0 and 1-\frac{1}{x}. When added them all, I got: \frac{1}{x^2}<\sum _{k=1}^n a_k^2 < \frac{n}{x^2} - \frac{2(n-1)}{x} + n - 1. But i didn't know what to do next. How can I prove from here that there is no solution for the inequality \sum_{k = 1}^{n} a_k^2 < \frac{1}{n}?

    Thank you
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Bacterius's Avatar
    Joined
    Nov 2009
    From
    Wellington
    Posts
    927
    EDIT : forget this stuff, I got it wrong. Hang on, I'm thinking of something ...

    Yeah, I think I got it. Say you got your numbers, all equal to each other (so they all equal \frac{1}{n}). If you add some value x to one number, you need to report the opposite of this value onto the other numbers to balance the equality. Now which values of x don't allow any balancing of the other numbers that satisfies the sum of the squares being less than \frac{1}{n} ? Try to express this in mathematical terms. You should find that x has no solution ... your problem follows. I might post a proof, I'm actually as stumped as you on this one
    Last edited by Bacterius; February 28th 2010 at 04:00 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member u2_wa's Avatar
    Joined
    Nov 2008
    Posts
    119
    Quote Originally Posted by mohammadfawaz View Post
    Hello all,

    I need help proving the following:
    if a,b,c,... are n real positive number such that a+b+c+... = 1, then
    a^2 + b^2 + c^2 + ... >=1/n

    Thank you
    Hello mohammadfawaz
    If a+b+c=1
    Let A=average of a+b+c
    we can say that a^2+b^2+c^2>=A^2+A^2+A^2
    Similarly,
    a+b+c+d+.....=1, A=\frac{1}{n}<br />

    a^2+b^2+c^2+....>=nA^2
    a^2+b^2+c^2+....>=\frac{1}{n}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member mohammadfawaz's Avatar
    Joined
    Feb 2010
    From
    Lebanon - Beirut
    Posts
    100
    hello u2_wa

    There is one small thing I didn't get in your proof:
    by saying a^2+b^2+c^2>=A^2+A^2+A^2, you're assuming that: a^2>=A^2 and b^2>=A^2 and c^2>=A^2. But if A is the average of a,b and c, then A may be smaller or larger than a, b or c. Would you please explain how did you get the inequality?

    Thank you
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member u2_wa's Avatar
    Joined
    Nov 2008
    Posts
    119
    Quote Originally Posted by mohammadfawaz View Post
    hello u2_wa

    There is one small thing I didn't get in your proof:
    by saying a^2+b^2+c^2>=A^2+A^2+A^2, you're assuming that: a^2>=A^2 and b^2>=A^2 and c^2>=A^2. But if A is the average of a,b and c, then A may be smaller or larger than a, b or c. Would you please explain how did you get the inequality?

    Thank you
    Hello mohammadfawaz:

    Certainly I am not assuming a^2>=A^2 and b^2>=A^2 and c^2>=A^2

    I am saying that sum of a^2+b^2+c^2 will always be equal to or greater than
    A^2+A^2+A^2

    To prove a^2+b^2+c^2>=A^2+A^2+A^2

    a^2+b^2+c^2>=3A^2

    a^2+b^2+c^2>=3(\frac{a+b+c}{3})^2

    \frac{3a^2+3b^2+3c^2-(a+b+c)^2}{3}>=0

    \frac{(a-c)^2+(a-b)^2+(b-c)^2}{3}>=0 (PROOF)

    Same rule applies to a+b+c+d+.....=nA

    Hope this helps!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving Euler's Formula
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: November 9th 2011, 09:25 AM
  2. proving a formula
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 29th 2011, 02:23 AM
  3. Proving the reduction formula?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 21st 2011, 10:27 PM
  4. Proving variance formula
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 8th 2011, 05:25 PM
  5. Proving a formula ..
    Posted in the Calculus Forum
    Replies: 6
    Last Post: August 24th 2010, 11:05 AM

Search Tags


/mathhelpforum @mathhelpforum