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Showing $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$?

Discussion in 'Mathematics' started by user106860, Aug 1, 2020.

  1. user106860

    user106860 Guest

    I am wondering if we can Show that $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$?

    (I'm particularly interested in $0<a<b<1$ but I don't think restricting $a$ and $b$ here matters)

    I think I've seen an answer using polar coordinates, so perhaps that way could be used, but can we avoid polar coordinates too?

    So to state the question precisely:


    can we show that $a^2 + b^2 > 2ab$ without using the fact that $(a-b)^2 = a^2 + b^2 -2ab$, and without using polar coordinates?

    basically, I am wondering if there is something like (assuming $a<b$:

    $a^2 + b^2 > 2a^2$,

    and then somehow showing that $2a^2 > 2ab$

    I know that the above way cannot work, because $2a^2$ need not be greater than $2ab$, but perhaps there is some similar approach.


    Alternatively, an answer saying why and approach like the one above cannot work (if it cannot work).

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