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). Login To add answer/comment