Results 1 to 6 of 6

Math Help - Prove that

  1. #1
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    541

    Prove that

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Mar 2008
    Posts
    182
    Quote Originally Posted by dhiab View Post
    Let b = 1 - a.
    Gives:
    \sqrt { a^2 + 1 - a} + \sqrt{  1 - a + a^2} + \sqrt{ 1 + a - a^2}

    0 <= a <= 1
    so the max value that any of the three square roots can have is 1.
    (you may wish to elaborate on this)

    And the sum of the three square roots must then be less than or equal to 3.
    Last edited by Unenlightened; July 29th 2009 at 05:03 AM. Reason: My inequality was unnecessarily strict...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by Unenlightened View Post
    0 <= a <= 1
    so the max value that any of the three square roots can have is 1.
    (you may wish to elaborate on this)
    The first two square roots are \le1 but the last one isn’t.

    0\le a\le1\ \implies\ \sqrt{1+a-a^2}\ge1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2008
    Posts
    182
    Ah feck. In future, I shall be more careful...

    Edit: In the words of H.H.Munro
    "A little inaccuracy sometimes saves a world of explanation."


    Right so, the third square root may be greater than 1, but for every increase above 1 in that, there is double the decrease in the sum of the other two square roots, so the greater the value of the third square root, the lower the value overall.
    I doubt that counts as a proof though..
    Last edited by Unenlightened; July 29th 2009 at 05:25 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    By the Generealised-Mean Inequality (*): <br />
\left( {\tfrac{{\sqrt {a^2  + b}  + \sqrt {b^2  + a}  + \sqrt {1 + ab} }}<br />
{3}} \right)^2  \leqslant \left( {\tfrac{{\left( {a^2  + b} \right) + \left( {b^2  + a} \right) + \left( {1 + ab} \right)}}<br />
{3}} \right)^1 <br />

    Now: <br />
\left( {a^2  + b} \right) + \left( {b^2  + a} \right) + \left( {1 + ab} \right) = \left( {a + b} \right)^2  + \left( {a + b} \right) + 1 - ab = 3 - ab \leqslant 3<br />

    Thus: <br />
\left( {\tfrac{{\sqrt {a^2  + b}  + \sqrt {b^2  + a}  + \sqrt {1 + ab} }}<br />
{3}} \right)^2  \leqslant 1<br />
from which we get: <br />
\sqrt {a^2  + b}  + \sqrt {b^2  + a}  + \sqrt {1 + ab}  \leqslant 3<br />
\square

    (*) See here
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    I think Cauchy–Schwarz might also work here.

    Note that the sum of the terms inside the square roots is a^2+ab+b^2+a+b+1=(a+b)^2+a+b+1-ab=3-ab\leq3.

    \therefore\ 1\cdot\sqrt{a^2+b}+1\cdot\sqrt{a+b^2}+1\cdot\sqrt{  1+ab}

    \le\ \sqrt{(1^2+1^2+1^2)(3-ab)}

    \le\ \sqrt{3\cdot3}

    =\ 3
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove a/b and a/c then a/ (3b-7c)
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: March 23rd 2010, 06:20 PM
  2. prove,,,
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: March 1st 2010, 10:02 AM
  3. Prove |w + z| <= |w| +|z|
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 28th 2010, 06:44 AM
  4. Replies: 2
    Last Post: August 28th 2009, 03:59 AM
  5. How to prove that n^2 + n + 2 is even??
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 30th 2008, 02:24 PM

Search Tags


/mathhelpforum @mathhelpforum