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  1. #1
    Member great_math's Avatar
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    Prove:

    Prove the following inequalities.

    1) $\displaystyle \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y}$

    2) $\displaystyle \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    What I have Done is WRONG as rightly pointed by Isomorphism

    Quote Originally Posted by great_math View Post
    Prove the following inequalities.
    1) $\displaystyle \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y}$

    2) $\displaystyle \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$
    1)


    Here$\displaystyle [\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]^2$

    $\displaystyle
    =\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} +2(1+1+1)
    $
    Here first term ie;
    $\displaystyle
    \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2}

    $
    is greater than zero and so is second term
    and
    $\displaystyle
    [\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]^2 \ge[\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]
    $
    If still in troble feel free to ask
    Last edited by ADARSH; Jan 10th 2009 at 05:55 AM.
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  3. #3
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by great_math View Post
    Prove the following inequalities.

    1) $\displaystyle \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{z} + \frac{y}{x} + \frac{z}{y}$
    Call $\displaystyle \frac{x}{y} = a, \frac{y}{z} = b , \frac{z}{x} = c$. Then observe that your inequality amounts to proving $\displaystyle a^2 + b^2 + c^2 \geq ab + bc + ca$ But $\displaystyle a^2 + b^2 + c^2 \geq ab + bc + ca \Leftrightarrow \frac12 ((a-b)^2 + (b-c)^2 + (c-a)^2) \geq 0$, which is clearly true.


    Quote Originally Posted by great_math View Post
    2) $\displaystyle \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \ge \frac{x}{y} + \frac{y}{z} + \frac{z}{x}$
    EDIT: This proof is wrong
    ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
    Multiply out throughout by $\displaystyle x^2y^2z^2$ and then the inequality reduces to $\displaystyle x^4z^2 + z^4y^2 + y^4 x^2 \geq x^3yz^2 + y^3zx^2 + z^3 x y^2$, which is clearly true by Muirhead inequality since (4,2,0) majorises (3,2,1).
    ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

    Quote Originally Posted by ADARSH View Post

    1)
    Here$\displaystyle [\frac{x}{z} + \frac{y}{x} + \frac{z}{y}]^2$

    $\displaystyle
    =\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} +2(1+1+1)
    $
    Thats wrong Adarsh, $\displaystyle \left(\frac{x}{z} + \frac{y}{x} + \frac{z}{y}\right)^2 =\frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} +2\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)
    $
    Last edited by Isomorphism; Jan 21st 2009 at 12:59 AM. Reason: Thanks Jane, Sorry great_math
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  4. #4
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Isomorphism View Post
    $\displaystyle x^4z^2 + z^4y^2 + y^4 x^2 \geq x^3yz^2 + y^3zx^2 + z^3 x y^2$, which is clearly true by Muirhead inequality since (4,2,0) majorises (3,2,1).
    Doesn’t Muirhead give $\displaystyle x^4y^2 + x^4z^2 + y^4x^2 + y^4z^2 + z^4x^2 + z^4y^2 \geq x^3y^2z + x^3z^2y + y^3x^2z + y^3z^2x + z^3x^2y + z^3y^2x$ instead?
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