Find the maximum of the function $f(x, y, z)=3 x+5 y-z$ on the sphere of radius 1, and the points where this maximum is attained.
Let's recall that a point is on the sphere of radius 1 if and only if $x^{2}+y^{2}+z^{2}=1$. This strongly resembles a term in the Cauchy-Schwarz inequality, let's make it appear:

$$
(3 x+5 y-z) \leq \sqrt{\left(x^{2}+y^{2}+z^{2}\right)\left(3^{2}+5^{2}+(-1)^{2}\right)}=\sqrt{14}
$$

To show that this is indeed the maximum, we need to find a point for which this becomes an equality. The theorem states that equality in Cauchy-Schwarz holds if and only if the families are proportional. Therefore, we need to find $(x, y, z)$ proportional to $(3,5,-1)$ and that satisfies $x^{2}+y^{2}+z^{2}=1$. There are two such points $\pm\left(\frac{3}{\sqrt{14}}, \frac{5}{\sqrt{14}}, \frac{-1}{\sqrt{14}}\right)$. The point with a + works (the other one achieves the minimum $-\sqrt{14})$.