<s>[INST] Real Analysis Inequality Questions I am working on a few questions, and I was hoping for some feedback regarding whether or not my answers make sense. 

(a) Let $\mathbf{x}, \mathbf{y}, \mathbf{z} \in \mathbb{R}^d$. If $||\mathbf{x}||<2$, $||\mathbf{y}||<3$ and $||\mathbf{z}||<4$, show that 
  $$| \mathbf{x} \cdot \mathbf{y} - \mathbf{x} \cdot \mathbf{z}| < 14.$$ 

Observe that $\mathbf{x} \cdot \mathbf{y}=|| \mathbf{x} |||| \mathbf{y}|| cos (\theta)$, where $\theta $ is the angle between $\mathbf{x}$ and $\mathbf{y}$, and $\mathbf{x} \cdot \mathbf{z}=|| \mathbf{x} |||| \mathbf{z}|| cos (\alpha)$, where $\alpha $ is the angle between $\mathbf{x}$ and $\mathbf{z}$. Then we have 
$$\mathbf{x} \cdot \mathbf{y} - \mathbf{x} \cdot \mathbf{z} = || \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha).$$
The function $cos(x)$ ranges between $-1$ and $1$. Now, I believe that the extreme values of the expression above are $-14$ and $14$. Note that $|| \mathbf{x} |||| \mathbf{y}||<6$ and $|| \mathbf{x} |||| \mathbf{z}|| < 8$. Furthermore, $ -1 \leq cos(\theta) \leq 1$ and $ -1 \leq cos(\alpha) \leq 1$. If $cos(\theta)=1$, and $cos(\alpha)=-1$, then 
$$|| \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha) = || \mathbf{x} |||| \mathbf{y}|| + || \mathbf{x} |||| \mathbf{z}|| < 6 +8=14.$$
If $cos(\theta)=-1$, and $cos(\alpha)=1$, then we multiply the previous inequality on each sides by $(-1)$, and change the direction of the inequality accordingly, to arrive at 
$$- || \mathbf{x} |||| \mathbf{y}|| - || \mathbf{x} |||| \mathbf{z}|| > -6 -8=-14.$$
This gives, 
$$-14 <|| \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha) < 14$$
or,
$$\left |  || \mathbf{x} |||| \mathbf{y}|| cos (\theta) - || \mathbf{x} |||| \mathbf{z}|| cos (\alpha) \right | < 14,$$
$$| \mathbf{x} \cdot \mathbf{y} - \mathbf{x} \cdot \mathbf{z}| < 14.$$
Is this correct? If it is, is my explanation of the extreme values rigorous enough? 

(b) Set $B:= \left \{ \mathbf{x} \in \mathbb{R}^n : ||\mathbf{x}|| <1  \right \}.$ If $\mathbf{x}, \mathbf{y}, \mathbf{z} \in B$, show that
  $$\mathbf{w} = \frac{(\mathbf{x} \cdot \mathbf{y})\mathbf{z} + (\mathbf{x} \cdot \mathbf{z})\mathbf{y} + (\mathbf{z} \cdot \mathbf{y})\mathbf{x}}{3}$$
  belongs to B.

Using the same reasoning as before, and defining the dot products as the product of the lengths of the vectors and the cosine of the angle between the vectors, I get
$$|| \mathbf{w} || = \left \| \frac{(\mathbf{x} \cdot \mathbf{y})\mathbf{z} + (\mathbf{x} \cdot \mathbf{z})\mathbf{y} + (\mathbf{z} \cdot \mathbf{y})\mathbf{x}}{3} \right \| < \left \| \frac{\mathbf{z} + \mathbf{y} + \mathbf{x}}{3} \right \| < \frac{||\mathbf{z}||}{3} + \frac{||\mathbf{y}||}{3} + \frac{||\mathbf{x}||}{3} < 1$$
So, $\mathbf{w} \in B$.

(c) Let B be as above, and let $ \mathbf{z}$, $\mathbf{w} \in B$. Prove that
  $$|\mathbf{x} \cdot \mathbf{z} - \mathbf{y} \cdot \mathbf{w} | \leq \left \| \mathbf{y}-\mathbf{z} \right \| + \left \| \mathbf{x}-\mathbf{w} \right \|$$

$$|\mathbf{x} \cdot \mathbf{z} - \mathbf{y} \cdot \mathbf{w} | \leq |\mathbf{x} \cdot \mathbf{z} - \mathbf{z} \cdot \mathbf{w} + \mathbf{z} \cdot \mathbf{w} - \mathbf{y} \cdot \mathbf{w} | $$
$$\leq |\mathbf{z}| \cdot |\mathbf{x} - \mathbf{w}| + |\mathbf{w}| \cdot |\mathbf{z} - \mathbf{y}|$$
$$\leq ||\mathbf{z}|| \cdot ||\mathbf{x} - \mathbf{w}|| + ||\mathbf{w}|| \cdot ||\mathbf{z} - \mathbf{y}||$$
$$\leq ||\mathbf{x} - \mathbf{w}|| +  ||\mathbf{z} - \mathbf{y}||$$
Any input would be appreciated.
 [/INST] a) follows from $|x \cdot y - x \cdot z| = |x \cdot (y -z)| \leq \|x\| \|y-z\| \leq \|x\| (\|y\|+\|z\|) = 2 (3+4)$.
The first inequality in your answer for b) is incorrect. First do the triangle inequality, then upper bound using the fact that $|x \cdot y| < 1$ when $x,y \in B$.
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