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Let $f:[\alpha, \beta] \to \mathbb{R}^n$ be a smooth curve such that $f(\alpha)=\vec u$ and $f(\beta)=\vec v$. I want to show that for any unit vector $\vec n$, $$(\vec v - \vec u ) \cdot \vec n \leq \int_\alpha^{\beta} \left \lVert f'(t) \right \rVert \, dt$$

I'm aware that for any unit vector $\vec n$ we have that $\vec w \cdot \vec n \leq \left \lVert \vec w \right \rVert$ and I assume this is useful here, I'm not sure exactly how though.

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In $\mathbb{R}^n$ (with the Euclidean norm) the shortest path between two vectors $\vec u$ and $\vec v$ is the straight segment from $\vec u$ to $\vec v$, which has length $||\vec v-\vec u||$. This tells you that $$ ||\vec v-\vec u||\leq\text{length}\,(f)=\int_\alpha^\beta||f'(t)||\,dt, $$ which along with your observation gives the desired result.

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    The proof is good, just a small note: it holds for all norms derived from an inner product (use [Cauchy-Schwarz](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality) on the vectors $f'(t)$ and $f(\beta)-f(\alpha)$ and integrate over $[\alpha,\beta]$). – 2017-01-24
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    @Winther After first reading the question I suspected the result to be true with more generality, and actually (and coincidentally) the first thing that popped into my head was the Cauchy-Schwarz inequality, but I didn't see a way to use it and prove the claim. Thank you for the note! :) – 2017-01-25
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    @mathbeing How do you use the Cauchy-Schwarz inequality to prove the claim? – 2017-01-25
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    @Si.0788 Just follow Winther's hint!! It is pretty straighforward. – 2017-01-26