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Suppose $\vec x, \vec y \in \mathbb{R}^n$ with $||\vec y||=1$. Show that $\vec x \cdot \vec y \leq ||\vec x||$.

From the Cauchy-Schwarz inequality it follows that $$\vec x\cdot \vec y \leq |\vec x \cdot \vec y| \leq ||\vec x || \, ||\vec y||=|| \vec x ||$$

Is this correct?

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    Yes. this is correct. – 2017-02-02
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    yes, it's correct – 2017-02-02
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    Let me just add that you get the same from $\vec{x}\cdot\vec{y}=||\vec{x}||\,||\vec{y}||\cos{\theta}$. – 2017-02-02

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