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).