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

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..

• What have you tried so far? What are your thoughts? Questions are better received and we can better help you here on MSE when you include your current progress. – jgon Nov 7 '18 at 17:01
• I have tried Cauchy-Schwartz inequality, |<x,y>| <= ||x|| •||y||, where <x,y> is dot product between x and y. – Hubert Mark Nov 7 '18 at 17:08
• Also triangle inequality, ||x + y || <= ||x|| + ||y|| – Hubert Mark Nov 7 '18 at 17:10
• Also distance formula, d(a,b) <= d(a,c) + d(c,a), where d(a,b) = || a - b|| – Hubert Mark Nov 7 '18 at 17:11
• There is this formula: Ilz - xII^2 + Ilz - yII^2 - Ilx - yII^2 =2<z - x, z - y> Maybe it helps. – Hubert Mark Nov 7 '18 at 17:19

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 $$\begin{split} \|x-y \|\cdot\|z\|& =\left\| \frac{\|y\|}{\|x\|} x - \frac{\|x\|}{\|y\|} y \right \|\|z\| \\ &= \left\| \frac{\|y\|\|z\|}{\|x\|} x - \frac{\|x\|\|y\|}{\|z\|}z + \frac{\|x\|\|y\|}{\|z\|}z -\frac{\|x\|\|z\|}{\|y\|} y \right \| \\ &\le \left\| \frac{\|y\|\|z\|}{\|x\|} x - \frac{\|x\|\|y\|}{\|z\|}z \right \| + \left\| \frac{\|x\|\|y\|}{\|z\|}z -\frac{\|x\|\|z\|}{\|y\|} y \right \| \\ &= \|y\|\left\| \frac{\|z\|}{\|x\|} x - \frac{\|x\|}{\|z\|}z \right \| + \|x\| \left\| \frac{\|y\|}{\|z\|}z -\frac{\|z\|}{\|y\|} y \right \|\\ &= \|y\|\cdot \|x-z\| + \|x\| \cdot \|y-z\|. \end{split}$$

• I just want to point out that the problem is known as Ptolemy inequality and your proof mimicks the usual proof of it using inversion. – timon92 Nov 8 '18 at 17:25
• @timon92 I have never heard about this inequality. Seems like this proof is proof no. 8 at this site ckrao.wordpress.com/2015/05/24/… – daw Nov 8 '18 at 19:21
• Following reference [2] for proof no. 8 in the blog entry [which for some reason isn't displaying properly in Firefox 52.9.0 on my ancient PC - anyone else having trouble reading it?], I find page 185 of Werner H. Greub, Linear Algebra (third edition 1967) to be very reassuring! – Calum Gilhooley Nov 8 '18 at 22:42

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.

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):

• OK, you are right. +1 – user1551 Nov 8 '18 at 17:51

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}