Proof $|x+y|+|x-y| \geq |x| + |y|$

Question

Proof that for all $x,y \in \mathbb{R}$: $$|x+y|+|x-y| \geq |x| + |y|$$

My (failed) attempt: $$|x + y| \leq |x| + |y|$$ $$|x - y| \leq |x| + |y|$$ $$|x + y|+|x - y| \leq 2(|x| + |y|)$$

Thank you

Answer 1

Note that $|a|+|b|\geqslant |a\pm b|$, so we have $$ \Big|\frac{x+y}{2}\Big|+\Big|\frac{x-y}{2}\Big|\geqslant \Big|\frac{x+y}{2}+\frac{x-y}{2}\Big|=|x|. $$ Similarly, $$ \Big|\frac{x+y}{2}\Big|+\Big|\frac{x-y}{2}\Big|\geqslant \Big|\frac{x+y}{2}-\frac{x-y}{2}\Big|=|y|. $$ Add them up, we get the desired inequality.

Answer 2

Hint: break up the problem into four cases (when each of $x,y$ is positive or negative).

Answer 3

We have,
$$\color{red}{(|x+y|+|x-y|)^2 = 2(x^2+y^2) +2|x^2-y^2|}$$ Whence,

\begin{align}&2|x^2-y^2| \geq -x^2-y^2+2|x| |y|= -(|x|-|y|)^2\\ &\Longleftrightarrow(x^2+y^2) +2|x^2-y^2|\geq 2|x||y|\\ &\Longleftrightarrow 2(x^2+y^2) +2|x^2-y^2|\geq x^2+y^2+2|x||y|\\ &\Longleftrightarrow (|x+y|+|x-y|)^2 \geq (|x| + |y|)^2\\ &\Longleftrightarrow|x+y|+|x-y| \geq |x| + |y| \end{align}

Answer 4

squaring both sides we get $$x^2+y^2+x^2+y^2+2|x+y||x-y|\geq x^2+y^2+2|x||y|$$ this is true since we have $$(|x|-|y|)^2+2|x+y||x-y|\geq 0$$