Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Do you think this inequality is correct? I try to prove it, but I cannot. Please hep me. Assume that $\|X\| < \|Y\|$, where $\|X\|, \|Y\|\in (0,1)$ and $\|Z\| \gg \|X\|,\|Z\| \gg \|Y||$. prove that $$\|X+Z\|-\|Y+Z\| \leq \|X\|-\|Y\|$$ and if $Z$ is increased, the left hand side become smaller. I pick up some example and see that this inequality is correct but I cannot prove it. Thank you very much.

share|cite|improve this question

migrated from Jul 7 '13 at 18:41

This question came from our site for professional mathematicians.

I only need to prove with X,Y,Z∈R^2, I'm sorry for this inconvenience, because I'm an engineer. – Trinh Chien Jul 7 '13 at 7:01

The inequality is false as stated. Let

$$ \begin{align} X &= (0.5,0)\\ Y &= (-0.7,0)\\ Z &= (z,0), 1 \ll z \end{align}$$

This satisfies all the conditions given. We have that

$$ \|X + Z\| - \|Y + Z\| = z + 0.5 - (z - 0.7) = 1.2 \not\leq -0.2 = \|X\| - \|Y\| $$

From the Calculus point of view, in $n$ dimensions, we can write

$$ \|X\| = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} $$

We have that when $\|X\| \ll \|Z\|$, we can approximate $\|X+Z\| \approx \|Z\| + X\cdot \nabla(\|Z\|)$. Now, $\nabla(\|Z\|) = \frac{Z}{\|Z\|}$ by a direct computation, so we have that

$$ \|X + Z\| - \|Y + Z\| \approx (X-Y) \cdot \frac{Z}{\|Z\|} $$

From this formulation we see that even in the cases where $X,Y$ are infinitesimal the inequality you hoped for cannot hold true. However, the right hand side of this approximation can be controlled by Cauchy inequality to get (using that $Z / \|Z\|$ is a unit vector).

$$ (X-Y) \cdot \frac{Z}{\|Z\|} \leq \|X - Y\| $$

So perhaps what you are thinking about is the following corollary of the triangle inequality

Claim: If $X,Y,Z$ are vectors in $\mathbb{R}^n$, then $$ \|X + Z\| - \|Y + Z\| \leq \|X - Y \| $$

Proof: We write $$ X + Z = (X - Y) + (Y + Z) $$ so by the triangle inequality $$ \|X + Z\| = \|(X - Y) + (Y+Z)\| \leq \|X - Y\| + \|Y + Z\| $$ rearranging we get $$ \|X + Z\| - \|Y + Z\| \leq \|X - Y\| $$ as desired.

Remark: if we re-write the expression using $-Z$ instead of $Z$, the same claim is true in an arbitrary metric space: Let $(S,d)$ be a metric space. Let $x,y,z$ be elements of $S$. Then $$ d(x,z) - d(y,z) \leq d(x,y) $$.

share|cite|improve this answer
Sorry for this inconvenience, but Could I ask you a question,prof. Willie Wong? is this equation correct? ∥X+Z∥−∥Y+Z∥≈(X−Y).(Z/‖Z‖) As I see, ∥X+Z∥−∥Y+Z∥ is a number and (X−Y).(Z/∥Z∥) is a matrix? I'm sorry if it is a stupid question. – Hoping_VN Jul 8 '13 at 11:11
$Z$ is a vector. $(X-Y)$ is a vector. Their dot product $(X-Y)\cdot Z$ is a scalar. – Willie Wong Jul 8 '13 at 11:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.