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Thread: inequality

  1. #1
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    inequality

    hello,i tried to solve the following inequality but found difficulties i tried AM GM but didn't seem to work

    a,b,c are positive real numbers with a+b^+c^4=28
    prove that abc≤64
    any help or advice is highly appreciated

    by amgm i had 28/3≥cubicroot(ab^2c^4) but didn't know how to advance further i did some other attempts but i failed

    Last edited by aray; Aug 21st 2017 at 10:13 AM.
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  2. #2
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    Re: inequality

    I am not sure that this helps because it uses calculus.

    $d > 0,\ e > 0\, f > 0, \text { and } d + e + f = 28.$

    $z = def - y(28 - d - e - f) \implies$

    $\dfrac{\delta z}{\delta d} = ef - y \implies ef = y \text { if } \dfrac{\delta z}{\delta d} = 0.$

    $\dfrac{\delta z}{\delta e} = df - y \implies df = y \text { if } \dfrac{\delta z}{\delta e} = 0.$

    $\dfrac{\delta z}{\delta f} = de - y \implies de = y \text { if } \dfrac{\delta z}{\delta f} = 0.$

    $\dfrac{\delta z}{\delta y} = 28 - d - e - f \implies d + e + f = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

    $ef = y \implies y > 0 \text { and } f = \dfrac{y}{e}.$

    $df = y \implies d * \dfrac{y}{e} = y \implies \dfrac{d}{e} = 1 \implies d = e.$

    $de = y \text { and } f = \dfrac{y}{e} \implies f = \dfrac{de}{e} = d = e \implies$

    $d = e = f = \dfrac{28}{3}.$

    In other words, to maximize (d * e * f) subject to (d + e + f) = 28, d = e = f = 28/3. How you discover that with algebra, I am not sure, but I suppose you can confirm it with algebra.

    $\therefore a = \dfrac{28}{3} < 9.5,\ b = \sqrt{\dfrac{28}{3}} < 3.1 , \text { and } c = \sqrt[4]{\dfrac{28}{3}} < 1.8.$

    $\therefore abc < 9.5 * 3.1 * 1.8 = 53.01 \le 64.$
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  3. #3
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    Re: inequality

    Quote Originally Posted by JeffM View Post
    I am not sure that this helps because it uses calculus.

    $d > 0,\ e > 0\, f > 0, \text { and } d + e + f = 28.$

    $z = def - y(28 - d - e - f) \implies$

    $\dfrac{\delta z}{\delta d} = ef - y \implies ef = y \text { if } \dfrac{\delta z}{\delta d} = 0.$

    $\dfrac{\delta z}{\delta e} = df - y \implies df = y \text { if } \dfrac{\delta z}{\delta e} = 0.$

    $\dfrac{\delta z}{\delta f} = de - y \implies de = y \text { if } \dfrac{\delta z}{\delta f} = 0.$

    $\dfrac{\delta z}{\delta y} = 28 - d - e - f \implies d + e + f = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

    $ef = y \implies y > 0 \text { and } f = \dfrac{y}{e}.$

    $df = y \implies d * \dfrac{y}{e} = y \implies \dfrac{d}{e} = 1 \implies d = e.$

    $de = y \text { and } f = \dfrac{y}{e} \implies f = \dfrac{de}{e} = d = e \implies$

    $d = e = f = \dfrac{28}{3}.$

    In other words, to maximize (d * e * f) subject to (d + e + f) = 28, d = e = f = 28/3. How you discover that with algebra, I am not sure, but I suppose you can confirm it with algebra.

    $\therefore a = \dfrac{28}{3} < 9.5,\ b = \sqrt{\dfrac{28}{3}} < 3.1 , \text { and } c = \sqrt[4]{\dfrac{28}{3}} < 1.8.$

    $\therefore abc < 9.5 * 3.1 * 1.8 = 53.01 \le 64.$


    There is something wrong with the last part of this argument.

    if a=16, b=2\sqrt{2}, c=\sqrt{2} then a b c=64 giving the maximum of a b c

    so a b c cannot be less than 53.01
    Last edited by Idea; Aug 21st 2017 at 02:31 PM.
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  4. #4
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    Re: inequality

    You are correct. I think I better give it up.

    I need to be looking at $d + e^2 + f^4= 28$ as the constraint if dealing with $z = def.$

    I'll work on that.
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  5. #5
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    Re: inequality

    i found abc≤28/8c because

    (a+b^2+c^4)^2≥4a(b^2+c^4)≥8abc^2
    Last edited by aray; Aug 21st 2017 at 03:36 PM.
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  6. #6
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    Re: inequality

    $a > 0,\ b > 0\, c > 0, \text { and } a + b^2 + e^4 = 28.$

    $z = abc - y(28 - a - b^2 - c^2) \implies$

    $\dfrac{\delta z}{\delta a} = bc - y \implies bc = y \text { if } \dfrac{\delta z}{\delta a} = 0.$

    $\dfrac{\delta z}{\delta b} = ac - 2by \implies ac = 2by \text { if } \dfrac{\delta z}{\delta b} = 0.$

    $\dfrac{\delta z}{\delta c} = ab - 4c^3y \implies ab = 4c^3y \text { if } \dfrac{\delta z}{\delta c} = 0.$

    $\dfrac{\delta z}{\delta y} = 28 - a - b^2 - c^4 \implies a + b^2 + c^4 = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

    $bc = y \implies y > 0 \text { and } c = \dfrac{y}{b}.$

    $ac = 2by \implies a * \dfrac{y}{b} = 2by \implies a = 2b^2.$

    $ab = 4c^3y \implies 2b^3 = 4c^3 * bc \implies 0.5b^2 = c^4 \implies c = \sqrt[4]{0.5b^2}.$

    $\therefore 28 = a + b^2 + c^4 = 2b^2 + b^2 + 0.5b^2 = 3.5b^2 \implies$

    $0.5 * 7b^2 = 28 \implies 0.5b^2 = 4 \implies b^2 = 8 = 4 * 2 \implies b = 2\sqrt{2} \implies$

    $a = 2 * 8 = 16 \text { and } c = \sqrt[4]{0.5 * 8} = \sqrt{2}.$

    And $16 + 8 + 4 = 28 \text { and } 16 * 2\sqrt{2} * \sqrt{2} = 16 * 2 * 2 = 64.$

    $\therefore abc \le 64.$
    Last edited by JeffM; Aug 21st 2017 at 04:04 PM.
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  7. #7
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    Re: inequality

    thank you sir
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  8. #8
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    Re: inequality

    Quote Originally Posted by aray View Post
    thank you sir
    You are welcome. Sorry for initial mistake. Idea came to rescue.
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  9. #9
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    Re: inequality

    Quote Originally Posted by JeffM View Post
    $a > 0,\ b > 0\, c > 0, \text { and } a + b^2 + e^4 = 28.$

    $z = abc - y(28 - a - b^2 - c^2) \implies$

    $\color{red}{\dfrac{\delta z}{\delta a} = bc - y \implies bc = y \text { if } \dfrac{\delta z}{\delta a} = 0.}$

    $\color{red}{\dfrac{\delta z}{\delta b} = ac - 2by \implies ac = 2by \text { if } \dfrac{\delta z}{\delta b} = 0.}$

    $\color{red}{\dfrac{\delta z}{\delta c} = ab - 4c^3y \implies ab = 4c^3y \text { if } \dfrac{\delta z}{\delta c} = 0.}$

    $\dfrac{\delta z}{\delta y} = 28 - a - b^2 - c^4 \implies a + b^2 + c^4 = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

    $bc = y \implies y > 0 \text { and } c = \dfrac{y}{b}.$

    $ac = 2by \implies a * \dfrac{y}{b} = 2by \implies a = 2b^2.$

    $ab = 4c^3y \implies 2b^3 = 4c^3 * bc \implies 0.5b^2 = c^4 \implies c = \sqrt[4]{0.5b^2}.$

    $\therefore 28 = a + b^2 + c^4 = 2b^2 + b^2 + 0.5b^2 = 3.5b^2 \implies$

    $0.5 * 7b^2 = 28 \implies 0.5b^2 = 4 \implies b^2 = 8 = 4 * 2 \implies b = 2\sqrt{2} \implies$

    $a = 2 * 8 = 16 \text { and } c = \sqrt[4]{0.5 * 8} = \sqrt{2}.$

    And $16 + 8 + 4 = 28 \text { and } 16 * 2\sqrt{2} * \sqrt{2} = 16 * 2 * 2 = 64.$

    $\therefore abc \le 64.$
    For the three lines in red, it appears a negative sign was missed. It does not affect the end result (as the negatives seem to come in pairs, bringing the result back to positive).

    $\dfrac{\partial z}{\partial a} = bc+y \Longrightarrow bc = -y \text{ if } \dfrac{\partial z}{\partial a} = 0$

    $\dfrac{\partial z}{\partial b} = ac+2by \Longrightarrow ac = -2by \text{ if } \dfrac{\partial z}{\partial b} = 0$

    $\dfrac{\partial z}{\partial a} = ab+4c^3y \Longrightarrow ab = -4c^3y \text{ if } \dfrac{\partial z}{\partial c} = 0$

    $bc = -y \Longrightarrow y<0$ and $c=-\dfrac{y}{b}$.

    $ac=-2by \Longrightarrow -a\dfrac{y}{b} = -2by \Longrightarrow a=2b^2$

    $ab=-4c^3y \Longrightarrow 2b^3=-4c^3(-bc) \Longrightarrow 0.5b^2=c^4 \Longrightarrow c = \sqrt[4]{0.5b^2}$

    The rest follows the same.
    Last edited by SlipEternal; Aug 22nd 2017 at 07:52 AM.
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  10. #10
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    Re: inequality

    Quote Originally Posted by SlipEternal View Post
    For the three lines in red, it appears a negative sign was missed. It does not affect the end result (as the negatives seem to come in pairs, bringing the result back to positive).

    $\dfrac{\partial z}{\partial a} = bc+y \Longrightarrow bc = -y \text{ if } \dfrac{\partial z}{\partial a} = 0$

    $\dfrac{\partial z}{\partial b} = ac+2by \Longrightarrow ac = -2by \text{ if } \dfrac{\partial z}{\partial b} = 0$

    $\dfrac{\partial z}{\partial a} = ab+4c^3y \Longrightarrow ab = -4c^3y \text{ if } \dfrac{\partial z}{\partial c} = 0$

    $bc = -y \Longrightarrow y<0$ and $c=-\dfrac{y}{b}$.

    $ac=-2by \Longrightarrow -a\dfrac{y}{b} = -2by \Longrightarrow a=2b^2$

    $ab=-4c^3y \Longrightarrow 2b^3=-4c^3(-bc) \Longrightarrow 0.5b^2=c^4 \Longrightarrow c = \sqrt[4]{0.5b^2}$

    The rest follows the same.
    Umm

    I set up z = abc MINUS y * (28 and so on).

    As you say it makes no difference.
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  11. #11
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    Re: inequality

    Quote Originally Posted by JeffM View Post
    Umm

    I set up z = abc MINUS y * (28 and so on).

    As you say it makes no difference.
    Correct, $abc - y(28-a-b^2-c^4) = abc-28y \color{red}{+} ay \color{red}{+} b^2y \color{red}{+} c^4y$ because of the distributive property.
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  12. #12
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    Re: inequality

    We can also solve the problem using AM-GM

    \frac{\frac{a}{4}+\frac{a}{4}+\frac{a}{4}+\frac{a}  {4}+\frac{b^2}{2}+\frac{b^2}{2}+c^4}{7}\geq \sqrt[7]{\frac{a^4 b^4 c^4}{4^42^2}}
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