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Thread: Proving inequality with a, b

  1. #1
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    Proving inequality with a, b

    Prove that for any positive real numbers a, b this inequality is valid:
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  2. #2
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    Assuming a and b are positive numbers, then
    for the first inequality:
    $\displaystyle \text{ since }2t^4-5t^3+6t^2-5t+2=(t-1)^2(2t^2-t+2)\geq 0 \text{ holds for all }t>0$.
    $\displaystyle \text{ By letting }t=\sqrt{\frac{a}{b}},\text{ we have }
    5(\frac{a}{b}+1)\sqrt{\frac{a}{b}}\leq 2(\frac{a}{b})^2+6(\frac{a}{b})+2$,
    Both sides times $\displaystyle b^2$, gives the first inequality.

    for the second inequality:
    $\displaystyle \text{ since }a^2+b^2\geq 2ab,\text{ we have }$
    $\displaystyle 4a^2+12ab+4b^2\leq 5a^2+10ab+5b^2=5(a+b)^2$.
    Both sides divided by $\displaystyle 10(a+b)$ gives the second inequality.
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  3. #3
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    Please, I don't understend how have you come to this inequality?

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

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

    What is this inequality deduced from?$\displaystyle 5(\frac{a}{b}+1)\sqrt{\frac{a}{b}}\leq 2(\frac{a}{b})^2+6(\frac{a}{b})+2$?? .....
    ...
    .... I hope you will understand. please help me.
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  4. #4
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    $\displaystyle \sqrt{ab}\leq\frac{2(a^2+3ab+b^2)}{5(a+b)}\leq\fra c{a+b}{2}$

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

    right inequality..

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

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

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

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

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

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

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

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

    Next prove the left inequality...
    Last edited by Archie Meade; Jan 9th 2010 at 08:54 AM.
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  5. #5
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    Big thanx.
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  6. #6
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    Left inequality...

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

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

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

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

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

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

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

    inconclusive

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

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

    Difference is

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

    the ratio is

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

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

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

    The ratio is $\displaystyle \leq1$,

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

    the difference between $\displaystyle 2(a+b)$ and $\displaystyle 4\sqrt{ab}$
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