Minimizing $(a\cdot b)(b\cdot c)(c\cdot a)$

Question

When $a$, $b$, and $c$ are unit vectors, what is the minimum value of $(a\cdot b)(b\cdot c)(c\cdot a)$?

If the three vectors are coplanar and mutually separated by $120^\circ$, then we get a value of $(-\frac12)^3=-\frac18$. Replacing any vector with its negative produces another configuration with the same value. I conjecture that this is the minimum. How would I prove this?

Equivalently, I want to prove that: $$\|a\|^2\|b\|^2\|c\|^2+8(a\cdot b)(b\cdot c)(c\cdot a)\ge0$$for all (not necessarily unit) vectors, with equality iff the three vectors are coplanar and the pairwise angles are each either $120^\circ$ or $60^\circ$.

One obvious approach is to use Cauchy–Schwartz inequality, $-{\|a\|}{\|b\|}\le a\cdot b\le{\|a\|}{\|b\|}$. However, we only get equality on that lower bound when $a$ and $b$ are antiparallel.

Applying Cauchy–Schwartz to each of the three inequalities, we obtain: $$-\|a\|^2\|b\|^2\|c\|^2\le(a\cdot b)(b\cdot c)(c\cdot a)$$ For unit vectors, it tells us that we'd get a minimum value of $-1$ if the three vectors were mutually antiparallel. However, this is an impossible configuration. So Cauchy–Schwartz looks useless.

Is there a (hopefully elementary) way to prove the above inequality, and confirm that, when $\|a\|=\|b\|=\|c\|=1$, the minimum value of $(a\cdot b)(b\cdot c)(c\cdot a)$ is $\frac18$?

Answer 1

HINT

Indicating with $0\le\alpha,\beta,\gamma\le \pi$ the angles between each pair of the three vectors, the condition is equivalent to minimize

• $\cos \alpha \cos \beta \cos \gamma$

under the constraint

• $\alpha+\beta+\gamma=a$ with $0\le a\le 2\pi$

By Lagrange's multipliers we find

• $-\sin \alpha \cos \beta \cos \gamma=\lambda$
• $-\cos \alpha \sin \beta \cos \gamma=\lambda$
• $-\cos \alpha \cos \beta \sin \gamma=\lambda$

which implies that the minimum is reached when

• $\alpha=\beta=\gamma=\frac a 3$

thus the extreme value is

• $\cos^3 \frac a 3\implies \cos^3 \frac {2\pi} 3=-\frac18$

Answer 2

From the inner product identity (sometimes taken as well as a definition) follows $$(a\cdot b)=||a||||b||\cos\theta_1\\ (b\cdot c)=||b|||c||\cos\theta_2\\ (c\cdot a)=||c||||a||\cos\theta_3$$ where $\theta_1,\theta_2,\theta_3$ are the angles between the respective vectors. All the three vectors can be coplanar but need not to so it must be the case that $0\leqslant \theta_1+\theta_2+\theta_3\leqslant 2\pi$ (if coplanar then their sum is indeed $2\pi$). So your original inequality becomes $$||a||^2||b||^2||c||^2+8(a\cdot b)(b\cdot c)(c\cdot a)\geqslant 0\Leftrightarrow ||a||^2||b||^2||c||^2(1+8\cos\theta_1\cos\theta_2\cos\theta_3)\geqslant 0$$ Consider the minization problem $$\min_{\theta\in\mathbb{R}^3}f(\theta):=||a||^2||b||^2||c||^2(1+8\cos\theta_1\cos\theta_2\cos\theta_3)$$ subject to $$0\leqslant \theta_1+\theta_2+\theta_3\leqslant 2\pi$$ The Lagrangian define it to be $$L(\theta,\lambda,\mu):=f(\theta)+\lambda(2\pi-\theta_1-\theta_2-\theta_3)+\mu(-\theta_1-\theta_2-\theta_3)$$ First order conditions would imply $$f_{\theta_1}=f_{\theta_2}=f_{\theta_3}=\lambda+\mu$$ and slackness conditions $$\lambda(2\pi-\theta_1-\theta_2-\theta_3)=\mu(-\theta_1-\theta_2-\theta_3)=0$$ and $\lambda,\mu\geqslant 0$. From the first order condition equation one gets $$\tan\theta_1=\tan\theta_2=\tan\theta_3$$ You can consider all cases $\{\theta_1=\theta_2=\theta_3,\theta_1=\theta_2=\theta_3+\pi,\text{etc}...\}$. For instance when $\theta_1=\theta_2=\theta_3$ then clearly the miminium lies in $2\pi\geqslant\theta>\pi/2$ (in fact it is reached when $\theta=2\pi/3$. You need to rule out other cases if possible.