find the minimum value of $$S=\vec{a}\cdot\vec{b}+2 \vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}$$ if $|a|=1,|b|=2,|c|=3$
The solution is pretty easy: $$|2\vec{a} +2\vec{c}+\vec{b}|^2\ge 0$$ $$\iff 4+36+4+4S\ge 0$$ $$\iff S\ge -11$$ However if i want to solve this like this it becomes difficult
Let $p,q,r$ be the angles btw the vectors $(a,b),(b,c),(c,a)$ respectively.Then we have to minimise $$S=2\cos p+6\cos q+6\cos r$$ We cannot take $p=q=r=-\pi$ as that would mean all vectors make an angle of $-\pi$ with each other which is insane!
How would i go about solving from this method??
UpdateAs Calvin Lin pointed out My first meethod is wrong
You need the additional constraint that $ p + q + r \leq 2 \pi $ (where we take the angle in the range $ [ 0, \pi ] $).
This is because the angles at the vertex/origin sum up to at most $2 \pi$.
Furthermore, because the coefficients are positive, we will always want to increase $p/q/r$, so the minimum will occur when $ p + q + r = 2 \pi$.
At this point, we can use Lagrange Multipliers to calculate the minimum.