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Thread: Equality

  1. #1
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    Equality

    For non-zero a,b,c

    $\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$

    show also that equality holds if and only if $\displaystyle a,b,c\in\{-1,1\}$.

    How can I show this?
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    Quote Originally Posted by dwsmith View Post
    For non-zero a,b,c
    $\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$
    show also that equality holds if and only if $\displaystyle a,b,c\in\{-1,1\}$.
    That "only if" part is not true.
    $\displaystyle \left( {\forall x,y} \right)\left[ {x^2 + y^2 \geqslant 2xy} \right]$.
    If $\displaystyle t\ne 0\text{ let }x=t~\&~y=\frac{1}{t},~t=a,b,c$.
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by dwsmith View Post
    For non-zero a,b,c

    $\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$

    show also that equality holds if and only if $\displaystyle a,b,c\in\{-1,1\}$.

    How can I show this?
    $\displaystyle a^2+\frac{1}{a^2}\geq2$

    $\displaystyle b^2+\frac{1}{b^2}\geq2$

    $\displaystyle c^2+\frac{1}{c^2}\geq2$

    (Using AM-GM inequality three times)

    Adding,

    $\displaystyle a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c ^2}\geq 6$

    Each of the AM-GM inequalities becomes an equality if and only if $\displaystyle a^2=\frac{1}{a^2}$, $\displaystyle b^2=\frac{1}{b^2}$, $\displaystyle c^2=\frac{1}{c^2}$

    ie $\displaystyle a=\pm1$, $\displaystyle b=\pm1$, $\displaystyle c=\pm1$

    ie $\displaystyle a,b,c\in\{-1,1\}$
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