Summation Summation Summation

Let $x_{1} = 50 - \alpha$ and $x_{k} = \dfrac{\alpha}{2014}$ for $2 \le k \le 2015$ for some $\alpha \gt 0.$

Let $y_{2} = 169 - \alpha$ and $y_{1} = y_{k} = \dfrac{\alpha}{2014}$ for $3 \le k \le 2015.$

Let $z_{3} = 961 - \alpha$ and $z_{1} = z_{2} = z_{k} = \dfrac{\alpha}{2014}$ for $4 \le k \le 2015.$

Then the initial three sum conditions are satisfied, and we find that

$S = \displaystyle\sum_{k=1}^{2015} x_{k}y_{k}z_{k} = (50 + 169 + 961 - 3\alpha)(\dfrac{\alpha}{2014})^{2} + 2012(\dfrac{\alpha}{2014})^{3}.$

We can then make $S$ arbitrarily small by making $\alpha$ arbitrarily small, but since $\alpha$ can never actually be $0$ , (since $x_{k}, y_{k}, z_{k} \gt 0$ for all $k$ ), $S$ can never be zero. Thus there is no minimum.

Note that, since all the terms of $S$ exceed $0$ by the given conditions, there is no way for $S$ to equal $0$ . But by the above argument, if $S$ did have a minimum associated with some $\alpha$ value, then we could just choose a smaller $\alpha$ value in order to find a "smaller minimum", hence giving rise to a contradiction. Thus we can conclude that there is no minimum. We could get more formal here, but this is the general concept at work.