Most concise way for derivation of$~|x+y|\geq|x|-|y|~$

Question

I want to derive the following inequality.

$$\left|x+y\right|\geq\left|x\right|-\left|y\right|\tag{1}$$

My tries for it are as following.

$$|x+y|\leq|x|+|y|~~\leftarrow~~\text{I omit derivation of it here}\tag{2}$$

$$|x|-|y|\leq|x|+|y|~~\leftarrow~~\text{obvious}\tag{3}$$

I've been got stucked from here.

I need your wisdom.

Answer 1

$$|x|=|x+y-y|\tag{1}$$

$$=\left|(x+y)+(-y)\right|\leq|(x+y)|+|(-y)|\tag{2}$$

$$\therefore~~|x|\leq|(x+y)|+|y|\tag{3}$$

$$\therefore~~|x|-|y|\leq|x+y|~~\leftarrow~~\text{QED}\tag{4}$$

Answer 2

Here is another way to approach it: \begin{align*} xy \geq -|xy| & \Longleftrightarrow 2xy \geq -2|xy|\\\\ & \Longleftrightarrow x^{2} + 2xy + y^{2} \geq x^{2} - 2|xy| + y^{2}\\\\ & \Longleftrightarrow (x + y)^{2} \geq |x|^{2} - 2|x||y| + |y|^{2}\\\\ & \Longleftrightarrow |x + y|^{2} \geq (|x| -|y|)^{2}\\\\ & \Longleftrightarrow |x + y| \geq ||x| - |y|| \end{align*}

Once $||x| - |y|| \geq |x| - |y|$, the claimed result holds.

Hopefully this helps!

Answer 3

The posted question reduces to proving that

$$|a+b| \leq |a| + |b|.$$

One method is:

• $|a+b| = (a+b)$ or $|a+b| = -(a+b),$ by definition.
• $a \leq |a|, b \leq |b| \implies (a+b) \leq |a| + |b|.$
• $-a \leq |a|, -b \leq |b| \implies -(a+b) \leq |a| + |b|.$