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

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

## closed as off-topic by YiFan, Shailesh, Alex Provost, Leucippus, Eevee TrainerMar 20 at 6:17

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• Suppose it was strictly less than. Subtract $x^2$ from both sides and you have a contradiction. – Rocket Man Mar 19 at 14:54
• oh that was very straightforward. Thank you – Shay Mar 19 at 14:59

## 2 Answers

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 $$x,y \geq 0$$.

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:

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