Let $S = \textbf{x}\cdot\textbf{y}+\textbf{y} \cdot \textbf{z}+\textbf{z} \cdot \textbf{x}$. Show the minimum of $S$ over $\textbf{z}$ given that $\textbf{x}$ and $\textbf{y}$ are fixed and linearly independent is attained when $\textbf{z} = \lambda(\textbf{x}+\textbf{y})$. Show also that for this value of $\lambda$
i) $\lambda\leq-\frac{1}{2}$ for any choice of $\textbf{x}$ and $\textbf{y}$
ii) $\lambda = -1$ and $S=-\frac{3}{2}$ when $\textbf{x}\cdot\textbf{y} = -\frac{1}{2}$
Since $\textbf{x}$ and $\textbf{y}$ are fixed, so is $\textbf{x}\cdot\textbf{y}$, so we focus on $\textbf{y} \cdot \textbf{z}+\textbf{z} \cdot \textbf{x} = \textbf{z}(\textbf{x}+\textbf{y}) = \lvert \textbf{z}\rvert \lvert \textbf{x}+\textbf{y}\rvert \cos(\theta)$, which is minimised when $\cos(\theta) = -1$, i.e. when $\textbf{z}$ is in the opposite direction to $\textbf{x}+\textbf{y}$, so when $\textbf{z} = \lambda(\textbf{x}+\textbf{y})$ with $\lambda \leq 0$. I don't see where the $-\frac{1}{2}$ comes in, or how to do ii).