Easy or hard !!!

We note that

$\begin{aligned} (a+b+c)^2 & = \red{a^2 + b^2 + c^2} + 2(ab+bc+ca) \\ \implies ab + bc + ca & = \frac {\blue{(a+b+c)^2} - \red 1}2 & \small \blue{\text{By Cauchy-Schwarz inequality}} \\ & \le \frac {\blue{(1^2+1^2+1^2)(a^2+b^2+c^2)} - 1}2 \\ & = \frac {(3)(1)-1}2 = 1 \end{aligned}$

Therefore $\max(ab+bc+ca) = 1$ , when $a=b=c = \frac 1{\sqrt 3}$ .

We also note that $(a+b+c)^2 \ge 0$ , therefore $\min \left((a+b+c)^2 \right) = 0$ , when $(a,b,c) = \left(\pm \frac 1{\sqrt 6}, \pm \frac 1{\sqrt 6}, \mp \frac 2{\sqrt 6} \right)$ . Therefore,

$\min (ab+bc+ca) = \dfrac {\min \left((a+b+c)^2\right)-1}2 = \dfrac {0-1}2 = - \dfrac 12$

And $\max(ab+bc+ca) + \min(ab+bc+ca) = 1 - \dfrac 12 = \dfrac 12 = \boxed{0.5}$ .

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Reference: Cauchy-Schwarz inequality

Consider the symmetric matrix $A = \begin{pmatrix} 0 && 0.5 && 0.5 \\ 0.5 && 0 && 0.5\\ 0.5 && 0.5 && 0\end{pmatrix}.$ The problem then reduces to finding the maxima and minima of the quadratic form $\bm{x}^T A \bm{x}$ , where $||\bm{x}||_2 =1$ . By the Courant-Fischer Theorem , the maximum and the minimum value of the quadratic under this constraint is given by the maximum and minimum eigenvalue of $A$ respectively. It can be readily computed that $\lambda_{\max}=1, \lambda_{\min}=-0.5.$ Hence the answer is $1-0.5=0.5.$

We have $a^2+b^2+c^2\geq ab+bc+ca$ for all real $a, b, c$ . The equality holds when $a=b=c$ .

For $a^2+b^2+c^2=1, ab+bc+ca\leq 1$ , the equality holds when $a=b=c=\pm \dfrac {1}{\sqrt 3}$

We also have $(a+b+c)^2\geq 0$ for all real $a, b, c$ , so that the minimum value of it is zero when $a+b+c=0$

So we need to find the values of $a, b, c$ satisfying $a+b+c=0$ subject to the constraint $a^2+b^2+c^2=1$

The solution is $a=\dfrac {-b\pm \sqrt {2-3b^2}}{2},c=-\dfrac {b\pm \sqrt {2-3b^2}}{2},-\sqrt {\frac{2}{3}}\leq b\leq \sqrt {\frac{2}{3}}$

So for all values of $b$ within the given interval and the corresponding values of $a$ and $c,(a+b+c)^2=0,a^2+b^2+c^2=1$ , so that $a^2+b^2+c^2=-2(ab+bc+ca)$

$\implies ab+bc+ca=-0.5$

So the maximum value of $ab+bc+ca$ is $1$ when $a=b=c=\pm \dfrac {1}{\sqrt 3}$ , and the minimum value is $-0.5$ when

$a=\dfrac {-b\pm \sqrt {2-3b^2}}{2},c=-\dfrac {b\pm \sqrt {2-3b^2}}{2},-\sqrt {\frac{2}{3}}\leq b\leq \sqrt {\frac{2}{3}}$

Hence the required sum is $1-0.5=\boxed {0.5}$

$a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}(2a^2+2b^2+2c^2-2ab-2bc-2ca)=\frac{1}{2}[(a-b)^2+ (b-c)^2+(c-a)^2]\geq 0$

$a^2+b^2+c^2-ab-bc-ac \geq 0$

$1-ab-bc-ac \geq 0$

$\Rightarrow$ $ab+bc+ca\leq 1$

$maximum$ value of $ab+bc+ca= 1$

$(a+b+c)^2\geq 0$

$a^2+b^2+c^2+2ab+2bc+2ac\geq 0$

$1+2ab+2bc+2ca\geq0$

$\Rightarrow$ $ab+bc+ca\geq-\frac{1}{2}$ \

$minimum$ value of $ab+bc+ca= \frac{1}{2}$

So,the sum of $maximum$ and $minimum$ value of $ab+bc+ca$ is $\frac{1}{2}$