Using the relationship mentioned above, we see that
\[\sqrt{\left(\sqrt{x^2}-\sqrt{y^2}\right)^2}\le \sqrt{(x+y)^2} \le \sqrt{x^2}+\sqrt{y^2}\]
Because all three expressions in the inequalities are positive, we can square each expression, yielding
\[\left(\sqrt{x^2}-\sqrt{y^2}\right)^2\le (x+y)^2 \le \left(\sqrt{x^2}\right)^2+2\sqrt{x^2y^2}+\left(\sqrt{y^2}\right)^2\]
Now we expand the binomials and use the relationship \(\left(\sqrt{a}\right)^2=a\), giving us the inequalities:
\[x^2-2|xy|+y^2\le x^2+2xy+y^2 \le x^2+2|xy|+y^2\]
and now by subtraction and factoring out identical terms we have \(-|xy|\le xy \le |xy|\).
Because all the transformations we performed are also valid in reverse, both of the original inequalities are valid.
