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?

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    $\begingroup$ Very nicely done. All you have to do io finish is demonstrate one set of allowable values for which this bound is achieved, and presto you have the minimum. Not very hard to find $(a, b, c)=(\frac1{\sqrt2}, -\frac1{\sqrt2}, 0)$ works. $\endgroup$
    – Macavity
    Dec 23, 2019 at 8:23

2 Answers 2


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 $a + b + c = 0$.

  • $\begingroup$ If I'm not wrong we should be getting infinite $\left(a,b,c\right)$ as the intersection of a sphere and a plane would be a circle $\endgroup$
    – sai-kartik
    Dec 23, 2019 at 5:49
  • $\begingroup$ @sai-kartik Yes, it's a circle on this sphere. $\endgroup$
    – Zarrax
    Dec 23, 2019 at 6:14

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.


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