# Inequality with absolute values: is it correct?

I'm looking for this inequality and it's making me confused

$$\left|\frac{xy}{\sqrt{x^2+y^2}}\right|\leq \frac{|x||y|}{\|(x,y)\|} \leq \frac{\|(x,y)\|^2}{\|(x,y)\|}$$

a) shouldn't the first inequality be an equality? I thought that the absolute value of a product is exactly the product of the absolute values (but the absolute value of a sum is less or equal to the sum of the absolute values). Since the bottom value is kept exactly the same in the 2nd term, shouldn't we have an equality?

b) I also don't get the second inequality (maybe I need a bit of more development in how we get there). Why is ${|x||y|} \leq {\|(x,y)\|^2}$ ? How do we know ${|x||y|} \leq x^2 + y^2$?

Thanks!

b) Note that if $a$ and $b$ are non-negative real numbers, then $ab\leqslant 2ab\leqslant a^2+b^2$, since $a^2+b^2-2ab=(a-b)^2\geqslant 0$.
use the inequality $$\frac{x^2+y^2}{2}\geq |xy|$$