Please help me to prove this inequality: $|x|+|y|≥|x+y|$

Question

Please help me to prove the following inequality:

$|x|+|y|\geq|x+y|$

in which $x$ and $y$ are real numbers.

Any help or hint would be appreciated. Thanks :)

Answer 1

Hint: If $w$ and $z$ are real numbers, $w,z \in \mathbb{R}$, (and not sets, as implied by your elementary-set-theory tag) then there are two cases: Either $w$ and $z$ have the same sign (both negative or both non-negative) or they don't (one is negative and the other non-negative). If $w$ and $z$ have the same sign, what happens to $w+z$? What if they have opposite signs? What's the largest $|w+z|$ can be?

Hint: If $w$ and $z$ are complex numbers, $w,z\in \mathbb{C}$, then let $w=r_1 e^{i\theta_1}$, $z=r_2 e^{i\theta_2}$. What happens to $|w+z|$ if $\theta_1 \equiv \theta_2\ (\operatorname{mod}\ 2\pi)$? What if $\theta_1 \not \equiv \theta_2\ (\operatorname{mod}\ 2\pi)$? What is the largest $|w+z|$ can be?

Hint: If $w$ and $z$ are sets, as not-stated in your question, by implied by your tag, then two cases occur again. Case 1: suppose $w$ and $z$ share at least one element. Case 2: suppose $w$ and $z$ do not share any elements. What is the largest $|w+z|$ can be?

Hint: If $\mathfrak{w}$ and $\mathfrak{z}$ are ideals of a ring $R$, then what you say is not true.

Meta hint: You should include more information in your questions.

Answer 2

Considering $w,z$ are complex numbers,

$|w+z|^2=(w+z)(\bar{w}+\bar{z})=|w|^2+|z|^2+(w\bar{z}+z\bar{w})$

$(|w|+|z|)^2=|w|^2+|z|^2+2|w||z|$

As $|w+z|,(|w|+|z|)\ge 0$ so to prove $|w+z|\le|w|+|z|$ it will suffice to prove $|w+z|^2\le(|w|+|z|)^2$

So it will suffice to prove

$(w\bar{z}+z\bar{w})\le2|w||z|$

Let $w=x+iy,z=a+ib$

Then $w\bar{z}=2(xa+yb)$

and $|w||z|=\sqrt{(x^2+y^2)(a^2+b^2)}$

So we are now reduced to proving,

$\sqrt{(x^2+y^2)(a^2+b^2)}\ge(xa+yb)$ and it follows from cauchy schwarz inequality with equality occuring when $\frac{x}{a}=\frac{y}{b}=k\Rightarrow w=kz$

Answer 3

geometrical interpretation: sum of the lengths of any two sides of triangle is greater than the length of the remaining side and $|x+y|=|x|^2+|y|^2-2|x||y|\cos(\alpha)$ $\alpha$ is angle between $x$ and $y$

Answer 4

Another answer

so

$|z+w|\leq |z|+|w|$