# 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. ### user106860Guest

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).