Prove $\forall x,y \in \mathbb{R} \ $ $x^2+y^2 \geq x^2-y^2$

Question

I know this may sound obvious, but I was wondering if both $x, y$ are real numbers, then why is it that $$x^2+y^2\geq x^2-y^2.$$

Answer 1

Note that $$ x^2 + {y^2} \geq x^2 \geq x^2 - y^2 $$ because $y^2 \geq 0$. Also, we do not require that $x,y \geq 0$.

Answer 2

Note: $y^2\ge 0$, $x^2\ge 0$, $x,y$ real

$y^2 \ge -y^2;$

Adding x^2 to each side of the above inequality:

$x^2 +y^2 \ge x^2-y^2$.