# Minimize $ab+bc+ca$ under three second degree constraints

As stated in the title, my problem is quite simple.

Minimize $ab+bc+ca$ under these three constraints:

$$a^2+b^2=1$$$$b^2+c^2=2$$$$c^2+a^2=2$$

I can brute force it, with some intelligence of course, and calculate the value of only three possibilities. But is there any way to do it with even more intelligence?

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## 2 Answers

First step is that you can get the possible values of $a,b,c$ according to $$a^2+b^2=1$$$$b^2+c^2=2$$$$c^2+a^2=2$$ Which are $$a=\pm\sqrt{1/2}$$$$b=\pm\sqrt{1/2}$$$$c=\pm\sqrt{3/2}$$ Notice that among the three products, $ab, bc$, and $ca$, at most 2 of them are negative, because at least 2 numbers among $a,b$ and $c$ have the same sign (i.e., these 2 numbers are either both negative, or positive). So if we can maximize the absolute values of these 2 negative products, we get the minimum value of $ab+bc+ca$. Which will be $$a=\sqrt{1/2}$$$$b=\sqrt{1/2}$$$$c=-\sqrt{3/2}$$ and under this condition, $$ab+bc+ca=1/2-\sqrt{3}$$

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$$b^{2} = {1 \over 2}\,\bracks{\overbrace{\pars{b^{2} + c^{2}}}^{=\ 2}\ +\ \overbrace{\pars{b^{2} - c^{2}}}^{=\ -1}} = {1 \over 2}$$

$$a^{2} = 1 - b^{2} = {1 \over 2}\,, \qquad\qquad c^{2} = 2 - b^{2} = {3 \over 2}$$

Then, $$a_{\pm} = \pm\,{\sqrt{2\,} \over 2}\,, \qquad b_{\pm} = \pm\,{\sqrt{2\,} \over 2} \,, \qquad c_{\pm} = \pm\,{\sqrt{6\,} \over 2}$$

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