Question on an absolute value inequality

Question

Is it true that, for every $x,y \ge 0$, $|x-y|\le |x+y|$? My geometric intuition says yes, but I might be missing something. Thanks!

Answer 1

Yes, because $\lvert x-y\rvert=\lvert x+(-y)\rvert\leqslant\lvert x\rvert+\lvert -y\rvert=x+y$.

Answer 2

Say $x\ge y\ge 0$ $$|x-y| = x-y\le x\le x+y$$

Answer 3

Of course.... for the reasons listed in other answers.

Answer 4

Square it: $$(x-y)^2\le (x+y)^2 \iff 4xy\ge 0 \quad \checkmark$$

Answer 5

$x,y \ge 0$;

$x+y \ge (x+y) -2y=x-y;$

$x+y \ge (x+y)-2x =y-x;$

Hence $x+y \ge |x-y|$.