From $||\alpha'||\geqq\alpha'\cdot\mathbf u$, deduce $L(\alpha)\geqq d(\mathbf {p,q})$, where $L(\alpha)$ is the length of $\alpha$

Question

Let $\alpha: [a,b]\to\Bbb R^3$ be an arbitrary curve segment from $\mathbf p=\alpha(a)$ to $\mathbf q=\alpha(b)$.

Let $\mathbf {u=\frac{q-p}{||q-p||}}$, the unit vector from $\mathbf p$ to $\mathbf q$.

From $||\alpha'||\geqq\alpha'\cdot\mathbf u$, deduce $L(\alpha)\geqq d(\mathbf {p,q})$, where $L(\alpha)$ is the length of $\alpha$ and $d$ is Euclidean distance.

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Assuming by "Length of $\alpha$" the arclength is meant, then I take it that $$L(\alpha)=\int_a^b ||\alpha'||dt$$ and by Euclidean distance, I assume that $$d(\mathbf {p,q})=||\mathbf {q-p}||$$ Integrating $||\alpha'||\geqq\alpha'\cdot\mathbf u$ from $a$ to $b$ gives $L(\alpha)\geqq\int_a^b{(\alpha'\cdot\mathbf u)dt}=\int_a^b||\alpha'||\cos(\theta)dt$ which is obvious, and gets me nowhere.

On the other hand, $||\alpha'||\geqq\alpha'\cdot\mathbf u=||\alpha'||\cos\theta \implies 1\geqq\cos\theta$, which is also obvious and takes me nowhere.

Yet another approach $$L(\alpha)\geqq\int_a^b{(\alpha'\cdot\mathbf u)dt}=\int_a^b(\alpha'\cdot\mathbf{\frac{q-p}{||q-p||}})dt=\frac{1}{\mathbf{||q-p||}}\int_a^b(\alpha'\cdot(\mathbf{q-p}))dt$$ which seems to be going in the opposite direction.

Hints (or full answers if you'd like) are appreciated.

Answer 1

$$\begin{split} \int_a^b \alpha' \cdot \mathbf u &= \int_a^b (\alpha \cdot \mathbf u)' \\ &= \alpha (b) \cdot \mathbf u - \alpha (a) \cdot \mathbf u\\ &= (\mathbf q - \mathbf p) \cdot u = \| \mathbf q - \mathbf p\| \end{split}$$