Cauchy-Schwarz inequality via optimisation

Lagrange. Maximize $f(x) = \sum x_i y_i$ under $g(x) = \sum x_i^2 = 1$. The gradients are $\nabla f = y$ and $\nabla g = 2x$. The condition $\nabla f = \lambda \nabla g$ gives
$$y = 2\lambda x \implies x = \frac{y}{2\lambda}.$$
Imposing $\|x\| = 1$:
$$\left\| \frac{y}{2\lambda} \right\| = 1 \implies |2\lambda| = \|y\| \implies x = \pm \frac{y}{\|y\|}.$$

Value. At $x = y/\|y\|$, $f(x) = \frac{y \cdot y}{\|y\|} = \|y\|$. At $x = -y/\|y\|$, $f(x) = -\|y\|$. So the maximum is $\|y\|$, and the minimum is $-\|y\|$.

Conclusion. For all $x$ with $\|x\|=1$, we have $|x \cdot y| \le \|y\|$. For arbitrary nonzero $x$, setting $u = x/\|x\|$ gives
$$|x \cdot y| = \|x\| \, |u \cdot y| \le \|x\| \, \|y\|.$$