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


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

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:

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


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