How to minimize $ab + bc + ca$ given $a^2 + b^2 + c^2 = 1$?

Question

The question is to prove that $ab + bc + ca$ lies in between $-1$ and $1$, given that $a^2 + b^2 + c^2 = 1$.

I could prove the maxima by the following approach. I changed the coordinates to spherical coordinates:

$a = \cos A \\ b = \sin A \cos B \\ c = \sin A \sin B$

Using that $\cos X + \sin X$ always lies between $- \sqrt 2$ and $\sqrt 2$ I proved that $a + b + c$ lies between $- \sqrt 3$ and $\sqrt 3$.

Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ I could prove that $ab + bc + ca$ is maximum at $1$ but I can't prove the minima.

Answer 1

From what you've done, $ab+bc+ca = \dfrac{(a+b+c)^2 - (a^2+b^2+c^2)}{2} \geq \dfrac{0 - 1}{2} = \dfrac{-1}{2}$, and this is the minimum value you sought. The minimum occurs when $a+b+c = 0, a^2+b^2+c^2 = 1$. To solve for $a,b,c$ you only need to find one solution of the system of $2$ equations above, then you are done. Since there are $3$ variables and only $2$ equations, you can take $c = \dfrac{1}{2}, \Rightarrow a, b$ are the solutions of the equation: $4x^2+2x-1 = 0$, thus $a = x = \dfrac{-1+\sqrt{5}}{4}, \Rightarrow b = -\dfrac{1}{2} - a = \dfrac{-1-\sqrt{5}}{4}$ by Viete's theorem on quadratic equation.

Answer 2

Hint: $a^2+b^2+c^2+ab+bc+ca=\frac{1}{2}((a+b)^2+(b+c)^2+(c+a)^2)\geq 0$.

Answer 3

The powerful Rayleigh quotient method can solve this problem, and many similar ones.

Indeed, this issue is equivalent to the maxi/minimization of the (Rayleigh) quotient $q(x)=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{X^TAX}{X^TX}$ with $A=\dfrac{1}{2}\begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$. The general result is that such a quotient takes its values in the closed interval $[\lambda_{min},\lambda_{max}]$ where $\lambda_{min}$ and $\lambda_{max}$ are resp. the minimal and maximal eigenvalues of $A$ and values $(a,b,c)$ providing the min (resp. max) are given by the associated (normed) eigenvectors.

Easy computations show that the eigenvalues of $A$ are $-1/2$ (multiplicity 2) and $1$, with associated normed eigenvectors:

• for $\lambda=-1/2$: $(a,b,c)=\dfrac{1}{\sqrt{2}}(-1,0,1)$ and $(a,b,c)=\dfrac{1}{\sqrt{2}}(-1,1,0)$,

• for $\lambda=1$: $(a,b,c)=\dfrac{1}{\sqrt{3}}(1,1,1)$.

Thus the interval of values taken by $ab+bc+ca$ is in between $\lambda_{min}=-0.5$ and $\lambda_{max}=1$, these extreme values being reached for vectors $(a,b,c)$ as given above.

For more on the Rayleigh quotient method, see https://en.wikipedia.org/wiki/Rayleigh_quotient ; for statistical applications, have a look at the slides "Rayleigh quotient" in http://www.svcl.ucsd.edu/courses/ece271B-F09/

Besides this algebraic approach, there is a geometric interpretation: let $S_k$ be the surface with equation $xy+yz+zx=k$. The maximum and minimum values for $k$ correspond to the limit cases for which $S_k \cap B \neq \varnothing$ where $B$ denotes the unit sphere.