The following optimization problem arose in the context of a magnetostatics problem:
Given unit vectors $a,b,c\in \mathbb{R}^n$, show that $3(a\cdot b)(a\cdot c)-(b\cdot c) \leq 2.$
(Compare the case of $n=2$ with problem 6.21(c) of Griffith's E&M textbook.)
This can be proven by Lagrangian multipliers without too much difficulty. To sketch this: One may assume without loss of generality that $a=e_1$. Then the method of Lagrangian multipliers shows that the extrema should occur when either $b_1=c_1=0$ or $b_k=c_k=0$ for all $k\neq 1$. In the former case, the objective function becomes $-(b\cdot c)\leq 1$; in the latter, we have $b,c$ being (anti)parallel to $a$ and the objective function achieves $\pm 3\|b\|c|\ \mp \|b\|c\| \leq 2$ as the desired upper bound.
However, the above proof seems overly tedious. Is there a simpler derivation?
$$ 3(a \cdot b)(a \cdot c) - (b \cdot c) = (3(a \cdot b)a-b) \cdot c. $$ This will be maximised when $c$ is parallel to $3(a \cdot b)a-b$, i.e. $$ c = \frac{3(a \cdot b)a-b}{\lVert 3(a \cdot b)a-b \rVert}, $$ so the maximum is $\lVert 3(a \cdot b)a-b \rVert = \sqrt{3(a\cdot b)^2+1} \leq 2$ by Cauchy–Schwarz.