Results 1 to 6 of 6

Math Help - Proving inequality with a, b

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    9

    Proving inequality with a, b

    Prove that for any positive real numbers a, b this inequality is valid:
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Shanks's Avatar
    Joined
    Nov 2009
    From
    BeiJing
    Posts
    374
    Assuming a and b are positive numbers, then
    for the first inequality:
    \text{ since }2t^4-5t^3+6t^2-5t+2=(t-1)^2(2t^2-t+2)\geq 0 \text{ holds for all }t>0.
    \text{ By letting }t=\sqrt{\frac{a}{b}},\text{ we have }<br />
5(\frac{a}{b}+1)\sqrt{\frac{a}{b}}\leq 2(\frac{a}{b})^2+6(\frac{a}{b})+2,
    Both sides times b^2, gives the first inequality.

    for the second inequality:
    \text{ since }a^2+b^2\geq 2ab,\text{ we have }
    4a^2+12ab+4b^2\leq 5a^2+10ab+5b^2=5(a+b)^2.
    Both sides divided by 10(a+b) gives the second inequality.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Please, I don't understend how have you come to this inequality?

    By the way, do you mean first inequality as this ineq.:: \sqrt{ab}\leq \frac{2(a^2+3ab+b^2)}{5(a+b)}

    How have you found out that t is \sqrt{\frac{a}{b}}?

    What is this inequality deduced from? 5(\frac{a}{b}+1)\sqrt{\frac{a}{b}}\leq 2(\frac{a}{b})^2+6(\frac{a}{b})+2?? .....
    ...
    .... I hope you will understand. please help me.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    \sqrt{ab}\leq\frac{2(a^2+3ab+b^2)}{5(a+b)}\leq\fra  c{a+b}{2}

    2\sqrt{ab}\leq\frac{4(a^2+3ab+b^2)}{5(a+b)}\leq\ a+b

    right inequality..

    \frac{4(a+b)^2+4ab}{5(a+b)}\leq\ a+b

    4(a+b)^2+4ab\leq\ 5(a+b)^2

    4ab\leq\ (a+b)^2

    (2\sqrt{ab})^2\leq\ (a+b)^2

    2\sqrt{ab}\leq\ a+b

    a+b-2\sqrt{ab}\geq0

    Since (\sqrt{a}-\sqrt{b})^2\geq0

    then a+b-2\sqrt{ab}\geq0

    Next prove the left inequality...
    Last edited by Archie Meade; January 9th 2010 at 09:54 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2010
    Posts
    9
    Big thanx.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Left inequality...

    5\sqrt{ab}\leq\frac{2(a+b)^2+2ab}{a+b}

    5\sqrt{ab}\leq2(a+b)+\frac{2ab}{a+b}

    We know from the right inequality that a+b\geq2\sqrt{ab}

    Hence 2(a+b)\geq4\sqrt{ab}

    We are now asking if \frac{2ab}{a+b}\geq\sqrt{ab}

    \frac{2\sqrt{ab}\sqrt{ab}}{a+b}\geq\sqrt{ab}\ ?

    2\sqrt{ab}\geq(a+b)\ ? No..

    inconclusive

    We can say... 2(a+b)\ is\ \geq4\sqrt{ab}
    by 2 times the amount that
    a+b\ is\ \geq2\sqrt{ab}.

    We can examine the case for \sqrt{ab}>\frac{2\sqrt{ab}\sqrt{ab}}{a+b}

    Difference is

    (\sqrt{ab})\frac{a+b-2\sqrt{ab}}{a+b}

    the ratio is

    \frac{\sqrt{ab}}{a+b}(a+b-2\sqrt{ab})

    Since... a+b\ge2\sqrt{ab}

    then.... 1\geq\frac{\sqrt{ab}}{a+b}

    The ratio is \leq1,

    therefore \sqrt{ab}\geq\frac{2ab}{a+b} by a margin not greater than

    the difference between 2(a+b) and 4\sqrt{ab}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. proving an inequality
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 6th 2011, 03:44 PM
  2. Replies: 3
    Last Post: December 12th 2010, 02:16 PM
  3. Proving an inequality #2.
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 12th 2010, 05:56 AM
  4. Proving an Inequality
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 1st 2010, 08:08 AM
  5. Help with proving an inequality
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 9th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum