Minimum value of $ab+bc+ca$ depending on given constraints

Question

We have to minimize the function $ab+bc+ca$ on the given condition that $a^2+b^2+c^2=1$. I have tried the following approach (but I don't think I'm right ) $$\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right) $$ We know that all squares are greater than equal to 0 and hence: $\left(a+b+c\right)^2\geq0$ $$\therefore a^2+b^2+c^2+2\left(ab+bc+ca\right) \geq0$$ $$\Rightarrow 1+2\left(ab+bc+ca\right) \geq0$$ $$\Rightarrow 2\left(ab+bc+ca\right)\geq -1$$ $$\Rightarrow \left(ab+bc+ca\right)\geq -1/2$$ As we can see I have reached a definite answer but my question is since there is a given constraint on $a$,$b$ and $c$ we can't say for sure that this value will be greater than or equal to -1/2, or can we?

Answer 1

You're doing it right. Another way of looking at what you're doing: You have $$ab + bc + ac = {1 \over 2} (a + b + c)^2 - {1 \over 2}(a^2 + b^2 + c^2)$$ Under the constraint that $a^2 + b^2 + c^2$, you therefore have $$ab + bc + ac = {1 \over 2} (a + b + c)^2 - {1 \over 2}$$ You're trying to minimize this quantity. Since ${1 \over 2} (a + b + c)^2 \geq 0$, it's smallest when $a + b + c = 0$, in which case you have $$ab + bc + ac = -{1 \over 2}$$ This is the minimum possible value of $ab + bc + ac$, and it is achieved for any $(a,b,c)$ for which $a^2 + b^2 + c^2 = 1$ and $a + b + c = 0$. There are many such $(a,b,c)$ as they correspond to the intersection of the sphere $a^2 + b^2 + c^2 = 1$ and the plane through the origin with equation $a + b + c = 0$.

Answer 2

Your method is absolutely correct. You have used the constraint $a^2+b^2+c^2=1$ in the second line of your working so there is no problem.