Norm inequality $\| x - y \| \cdot \| z \| \leq \| x - z \| \cdot \| y \| + \| z - y \| \cdot \| x \|$

Question

Let $x$, $y$, $z$ be $3$ vectors in Euclidean space $V$.

$\| x \|$ is norm of $x$ (length)

How do you prove:

$$\| x - y \| \cdot \|z\| \leq \| x - z\| \cdot \| y \| + \| z - y\| \cdot \| x \|?$$

I have tried Cauchy-Schwartz inequality, | < x,y > | <= ||x|| •||y||, where is dot product between x and y, trying to move everything on one side and then proving that it is greater than zero. Also I used modules properties, |a + b| < = |a| + |b| and so on.

Also triangle inequality, ||x + y || <= ||x|| + ||y|| , trying to get the other side of inequality.

Considering that ||x|| is sqrt( < x,x > ) and < x , a+b > is < x ,a > + < x ,b > and that, norm is always greater than 0 and sqrt( a+b )<= sqrt(a) + sqrt(b) , if a and b are positive, I played with this properties without getting any result..

Also distance formula, d(a,b) <= d(a,c) + d(c,a), where d(a,b) = || a - b||

There is this formula: Ilz - xII^2 + Ilz - yII^2 - Ilx - yII^2 =2 Maybe it helps. I have tried to use it but without a result..

Answer 1

If $x$, $y$, or $z$ is the zero vector, then the inequality is trivial. If the dimension $n=\dim V$ is $1$, the problem is also trivial. From now on, assume that $n\geq 2$ and $x,y,z\neq 0$, so the angles between these vectors are well-defined.

Let $a=\Vert x\Vert$, $b=\Vert y\Vert$, and $c=\Vert z\Vert$. Suppose that $\alpha=\angle(y,z)$, $\beta=\angle(z,x)$, and $\gamma=\angle(x,y)$. Clearly, we have $$\alpha,\beta,\gamma\in[0,\pi]\ \wedge\ \alpha+\beta+\gamma\leq 2\pi.$$ We also have triangle inequalities among the angles: $$\beta+\gamma\geq \alpha,\ \gamma+\alpha\geq \beta,\ \wedge\ \alpha+\beta\geq \gamma.$$ In particular, we have $$\gamma\leq \min\{\alpha+\beta,2\pi-\alpha-\beta\}.$$ That is, $$\frac{\gamma}{2} \leq \min\left\{\frac{\alpha+\beta}{2},\pi-\frac{\alpha+\beta}{2}\right\}.$$ Because $0\leq \alpha,\beta,\gamma\leq 2\pi$, we get $$0\leq \sin\frac{\gamma}{2}\leq \sin\frac{\alpha+\beta}{2}.\tag{1}$$

Now, $$\Vert x-y \Vert\cdot\Vert z \Vert=c\sqrt{(a-b)^2+4ab\sin^2\frac{\gamma}{2}}=\sqrt{(ca-bc)^2+4(ca)(bc)\sin^2\frac{\gamma}{2}}.$$ By (1), we get $$\Vert x-y \Vert\cdot\Vert z \Vert\leq \sqrt{(ca-bc)^2+4(ca)(bc)\sin^2\frac{\alpha+\beta}{2}}.$$ We also have $$\Vert y-z \Vert\cdot\Vert x \Vert=a\sqrt{(b-c)^2+4bc\sin^2\frac{\alpha}{2}}=\sqrt{(ab-ca)^2+4(ab)(ca)\sin^2\frac{\alpha}{2}}$$ and $$\Vert z-x\Vert\cdot\Vert y\Vert = b\sqrt{(c-a)^2+4ca\sin^2\frac{\beta}{2}}=\sqrt{(bc-ab)^2+4(bc)(ab)\sin^2\frac{\beta}{2}}.$$ So, the required inequality is immediate, if we can show that \begin{align}\sqrt{(ca-bc)^2+4(ca)(bc)\sin^2\frac{\alpha+\beta}{2}}&\leq \sqrt{(ab-ca)^2+4(ab)(ca)\sin^2\frac{\alpha}{2}}\\&\hphantom{12345}+\sqrt{(bc-ab)^2+4(bc)(ab)\sin^2\frac{\beta}{2}}.\end{align} For convenience, let $p=bc$, $q=ca$, and $r=ab$. We are to show that \begin{align}\sqrt{(p-q)^2+4pq\sin^2\frac{\alpha+\beta}{2}}&\leq \sqrt{(q-r)^2+4qr\sin^2\frac{\alpha}{2}}+\sqrt{(r-p)^2+4rp\sin^2\frac{\beta}{2}}.\tag{2}\end{align}

Let $u,v,w\in\mathbb{R}^2$ be vectors with $\Vert u \Vert=p$, $\Vert v\Vert =q$, and $\Vert w\Vert =r$ such that $\angle(v,w)=\alpha$ and $\angle(w,u)=\beta$, so that $\angle(v,u)=\alpha+\beta$ if $\alpha+\beta\leq \pi$, or $\angle(v,u)=2\pi-\alpha-\beta$ if $\alpha+\beta\geq \pi$. Then, we have $$\Vert u-v \Vert \leq \Vert w-v\Vert + \Vert u-w\Vert\tag{3}$$ by the triangle inequality. The inequality (3) is precisely (2), and the claim is proved.

Answer 2

This is a condensed version of @Zvi's answer. The proof below would not have been possible without that answer.

They key observation is the following identity: $$ \left\| \frac{\|y\|}{\|x\|} x - \frac{\|x\|}{\|y\|} y \right \| = \|x-y\|, $$ which can be proven by squaring. Using this indentity, we get $$\begin{split} \|x-y \|\cdot\|z\|& =\left\| \frac{\|y\|}{\|x\|} x - \frac{\|x\|}{\|y\|} y \right \|\|z\| \\ &= \left\| \frac{\|y\|\|z\|}{\|x\|} x - \frac{\|x\|\|y\|}{\|z\|}z + \frac{\|x\|\|y\|}{\|z\|}z -\frac{\|x\|\|z\|}{\|y\|} y \right \| \\ &\le \left\| \frac{\|y\|\|z\|}{\|x\|} x - \frac{\|x\|\|y\|}{\|z\|}z \right \| + \left\| \frac{\|x\|\|y\|}{\|z\|}z -\frac{\|x\|\|z\|}{\|y\|} y \right \| \\ &= \|y\|\left\| \frac{\|z\|}{\|x\|} x - \frac{\|x\|}{\|z\|}z \right \| + \|x\| \left\| \frac{\|y\|}{\|z\|}z -\frac{\|z\|}{\|y\|} y \right \|\\ &= \|y\|\cdot \|x-z\| + \|x\| \cdot \|y-z\|. \end{split}$$

Answer 3

Here is a rewritten form of daw's answer. Note that for any two scalars $p$ and $q$ and any two vectors $u$ and $v$ whose norms are equal, we have $$ \|pu-qv\|_2=\|qu-pv\|_2\tag{$\ast$} $$ because there exists a linear isometry $Q$ such that $Qu=v$ and $Qv=u$ (see also A symmetry identity involving norms).

Now, let $x=au,\ y=bv$ and $z=cw$ for some scalars $a,b,c$ and some unit vectors $u,v,w$. Then \begin{aligned} \|x-y\|\|z\| &=\|\color{red}{a}cu-\color{red}{b}cv\|\\ &=\|\color{red}{b}cu-\color{red}{a}cv\|\ \text{ by $(\ast)$}\\ &\le\|b\color{red}{c}u-\color{red}{a}bw\| + \|a\color{red}{c}v-a\color{red}{b}w\|\\ &=\|\color{red}{a}bu-b\color{red}{c}w\| + \|a\color{red}{b}v-a\color{red}{c}w\|\ \text{ by $(\ast)$}\\ &=\|x-z\|\|y\|+\|y-z\|\|x\|. \end{aligned}

Answer 4

You wait all day for a rewrite, and then three of them come along at once! :)

The inequality is clearly true if any of $x, y, z$ is zero, so we can suppose that none of them are.

The inequality then becomes: $$ d(x, y) \leqslant d(x, z) + d(z, y) $$ where: $$ d(x, y) = \frac{\|x - y\|}{\|x\|\|y\|} \ (x \ne 0, \ y \ne 0), \text{ etc.} $$ But we can write:

$$ d(x, y) = \|Tx - Ty\|, $$

where $T$ is a kind of "inversion": $$ T \colon V \setminus \{0\} \to V \setminus \{0\}, \ x \mapsto \frac{x}{\|x\|^2}. $$ Proof. As daw says, it's just a matter of squaring: $$ d(x, y)^2 = \frac{\left\langle x - y, x - y \right\rangle}{\|x\|^2\|y\|^2} = \frac{1}{\|y\|^2} - \frac{2\left\langle x, y \right\rangle}{\|x\|^2\|y\|^2} + \frac{1}{\|x\|^2} = \|Ty\|^2 - 2\left\langle Tx, Ty \right\rangle + \|Tx\|^2, $$ i.e. $d(x, y)^2 = \|Tx - Ty\|^2$. $\square$

And now we have: $$d(x, y) = \|Tx - Ty\| \leqslant \|Tx - Tz\| + \|Tz - Ty\| = d(x, z) + d(z, y), $$ as required. $\square$

---

From p.185 of Werner H. Greub, Linear Algebra (3rd ed. 1967) (see comments on daw's answer):

Answer 5

There is a proof based on the Quaternions.

First note that $x,y,z$ belong to a four dimensional vector space, where we can find an orthonormal basis such that the last two coordinates of $x$ and $y$ are zero.

Now we can replace $x,y,z$ by their vector coordinates on this basis and use the canonical inner product in $\mathbb{R}^4$ to prove the result, since the norms of the required vectors are unaltered under this simplification.

Next, we can regard $x,y,z$ as elements of the quaternions, but recall that the last two coordinates of $x,y$ are zero. Hence $x,y\in\mathbb{C}$.

Finally, $|(x-y)z|=|xz-yz|=|xz-xy+xy-yz|\leq |xz-xy|+|xy-yz|$

Since $x,y$ are complex numbers, they commute, then $|(x-y)z|\leq |xz-xy|+|yx-yz|$. So $$|(x-y)z|\leq |x(z-y)|+|y(x-z)|.$$

Recall that $|ab|=|a||b|$ for every pair of quaternions $a,b$, which completes the proof.