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Math Help - Problem 28

  1. #1
    Grand Panjandrum
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    Problem 28

    Proposition 1: If x+y+z=1 then xy+yz+xz<1/2

    Q1. Prove Proposition 1 is true

    Q2. Prove Proposition 1 is false

    There is a Q3 for when Q1 and Q2 have been settled.

    RonL
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  2. #2
    MHF Contributor red_dog's Avatar
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    If x,y,z\in\mathbf{R} then
    1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow
    \Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow
    \displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}. So the proposition is true.
    If x,y,z\in\mathbf{C} then let x=i,y=-i,z=1.
    Then \displaystyle xy+xz+yz=1>\frac{1}{2}. So the proposition is false.
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  3. #3
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    Quote Originally Posted by red_dog View Post
    If x,y,z\in\mathbf{R} then
    1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow
    \Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow
    \displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}. So the proposition is true.
    If x,y,z\in\mathbf{C} then let x=i,y=-i,z=1.
    Then \displaystyle xy+xz+yz=1>\frac{1}{2}. So the proposition is false.

    Q3. For x,y,z \in \mathbb{R} is the inequality tight, if not can you find a tight version.

    RonL
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  4. #4
    MHF Contributor red_dog's Avatar
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    For x,y,z\in\mathbf{R} the inequality is not tight.
    We have x^2+y^2+z^2\geq xy+xz+yz\Rightarrow
    \Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow
    \Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}.
    The equality stands for x=y=z=\frac{1}{3}.
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    Bar0n janvdl's Avatar
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    Hehehe, looks like this guy knows what he's doing, eh CaptainBlack?
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    Forum Admin topsquark's Avatar
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    "tight"?

    -Dan
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by red_dog View Post
    For x,y,z\in\mathbf{R} the inequality is not tight.
    We have x^2+y^2+z^2\geq xy+xz+yz\Rightarrow
    \Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow
    \Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}.
    The equality stands for x=y=z=\frac{1}{3}.
    You need to fill in some of the detail so others can follow this more easily.

    RonL
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  8. #8
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    x^2+y^2+z^2 \ge xy + yz + zx results from AM-GM inequality.
    Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
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    Quote Originally Posted by mathisfun1 View Post
    x^2+y^2+z^2 \ge xy + yz + zx results from AM-GM inequality.
    Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
    Actually that is Cauchy-Swartz
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  10. #10
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    Quote Originally Posted by mathisfun1 View Post
    x^2+y^2+z^2 \ge xy + yz + zx results from AM-GM inequality.
    Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
    Show us how.

    I see how it follows from the Cauchy Scwartz inequality:

    <br />
| \bold{x} \cdot \bold{y} |\le \| \bold{x} \|\ \| \bold{y} \|<br />

    Then putting \bold{x}=(a,b,c) and \bold{y}=(b,c,a), with a, b, c \in \mathbb{R}, we have:

    <br />
ab + bc + ca \le |ab + bc + ca| \le \sqrt{a^2+b^2+c^2}\ \sqrt{b^2+c^2+a^2} = a^2+b^2+c^2 <br />

    RonL
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  11. #11
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    Quote Originally Posted by CaptainBlank View Post
    Show us how.
    In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.
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  12. #12
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    Quote Originally Posted by ThePerfectHacker View Post
    In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.
    May be, but he should still make it explicit.

    Perhaps we should have a Wiki page on inequalities and their derivation/proof?

    RonL
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  13. #13
    MHF Contributor red_dog's Avatar
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    The inequality x^2+y^2+z^2\geq xy+yz+zx can be proved like this:
    Multiplying with 2, the inequality is equivalent to
    2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow
    \Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow
    \Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0.
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  14. #14
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    According to AM-GM, \frac{x^2+y^2}{2} \ge xy. Do the same for the other pairs of variables and add to get the desired inequality.


    Credit must be given where credit is due -- I picked up this trick from the AoPS book Vol 2.
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