# Minimum of $\textbf{x}\cdot\textbf{y}+\textbf{y} \cdot \textbf{z}+\textbf{z} \cdot \textbf{x}$

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).

• Hint: consider what value of lambda minimizes S given the restricted form of z you have derived. – John Polcari Apr 7 at 15:52
• @JohnPolcari why can't $\lambda$ be as small as possible? – user112358 Apr 7 at 16:09
• Plug and chug and I believe you will find the result has a minimum value because of how the length of x+y and the dot product x*y interact. – John Polcari Apr 7 at 16:12
• Or, more likely, not an actual minimum, but at least some obvious limits (like lambda <= -1/2). – John Polcari Apr 7 at 17:57