Prove $|X+Y| \le |X| + |Y|$

Question

Prove that $|X+Y| \le |X| + |Y|$ for all real numbers. I have come up with the answer that:

$|x+y|^2 \le (|x|+|y|)^2$

$x^2+y^2+2xy \le x^2+y^2+2|x||y|$

But I'm not sure that this solution fulfills all negative numbers like it does for all positive. If anyone could give their opinion that'd be great.

Answer 1

Notice that for all $x \in \mathbb R$: $|x| \geq \pm x$.

Then $x+y \leq |x|+y \leq |x|+|y|$ if $x+y$ is positive.

And if $x+y$ is negative, then $-x-y \leq |x|-y \leq |x|+|y|$ if $x+y$ is negative.

Answer 2

Yes, your are allowed to square both sides since both sides are already positive so proving $|x+y|^2\leq (|x|+|y|)^2$ is equivalent to triangle equality which is what you're proving. Generally the way you write the proof is $|x+y|^2=x^2+2xy +y^2\leq x^2+y^2 +2|x||y|=(|x|+|y|)^2$. Here the key step is that $xy\leq |x||y|$ which is called the Cauchy-Schwartz Inequality. So, this is far more general that just $x,y\in \mathbb{R}$.

Answer 3

You are right, and now you have to prove that $xy\le|xy|=|x||y|$.

and this is true since, $\forall x \in \mathbb{R} \Rightarrow x\le |x|$.

To prove this fact note that : if $x\ge 0$ than $|x|=x$ and if $x<0$ than $-|x|=x<0<|x|$.