How to solve $\;x \cdot y \cdot z < x + y + z\;$ inequality? I'm currently watching a climbing competition. Competitors will take part in three disciplines (Lead, Speed and Bouldering). For each discipline, the competitors gain points corresponding to their rank. 1 point competition leader. 2 points for the second and so on. The final score is obtained by multiplying the score obtained on each individual discipline. The competitor with the lowest final score will end with the competition with the best ranking.
For example, in the final round, Fujii Kokoro (JPN) gained 2 pts for lead, 2 pts for boulder and 5 pts for speed, making a total of 20 points.
I wonder how multiplying the scores (let's call them $x$, $y$ and $z$) leads to a different ranking than simply adding them. But I don't know how to solve that simple inequality:
$$x \cdot y \cdot z < x + y + z, \forall x > 0, y > 0, z > 0,  (,,) ∈ ℕ^3 $$
I'm less interested in the result than in understanding how to approach such problems.
 A: Unfortunately, there's not much to solve in that inequality, but there are some rearrangements that offer insight. If you have
$$xyz < x+y+z$$
you can divide both sides by $xyz$ to get that this is equivalent inequality
$$1 < \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}$$
which tells you that this inequality means that at least one of the products among $xy$, $xz$ and $yz$ has to be small (at most $3$) for the inequality to hold.
You can also, from the first inequality, read off important information: the term $xyz$ on the left has degree $3$, meaning that if you doubled every value, you would octuple ($=2^3$) the product $xyz$. The term $x+y+z$ has degree $1$ so doubling every value only doubles the total. This also suggests that the only possible solutions are small.
If you're interested in solutions with positive integers, this intuition suffices: first, let's just find solutions where $x\leq y\leq z$, since every solution is a rearrangement of such a solution. If $x\geq 2$, we then have $xyz \geq 4z$, but $x+y+z \leq 3z$, so there are no such solutions. Thus $x=1$ is the only possibility. Then, if $x=1$ and $y\geq 3$, we would get $xyz \geq 3z$ but $x+y+z \leq 1 + 2z$ which again rules out any solution. Thus $y$ is either $1$ or $2$. If $y$ is $2$, we can simply reduce this to asking which $z$ satisfy
$$2z < 1+2 +z$$
which reduces to $z < 3$. Then if $y=1$, we can reduce to $z < 1+1+z$, which is true for all $z$.
Thus, the set of ordered solutions to your inequality is $(1,1,z)$ and $(1,2,2)$.

This said, the inequality is not the best way to understand the difference. A better way is to understand the following identity:
$$\log(a)+\log(b) = \log(ab)$$
which converts between addition and multiplication. Saying that the winner is the person who minimizes $xyz$ is the same as asking that they minimize $\log(x)+\log(y)+\log(z)=\log(xyz)$. Using this correspondence, one can assign a point value to each place, and then just sum those points. Using a base $2$ logarithm, for instance, this assigns $0$ points to first place, $1$ point to second place, $1.58$ points to third, $2$ points to fourth, $2.32$ points to fifth, and $2.58$ points to sixth. Essentially, this creates a big bonus for first place over second place (equivalent to the bonus between second and fourth place or fourth and eighth*) - so your exact position is less important if you did badly in this scoring method than if positions were simply summed.
(*More precisely, the difference between $n^{th}$ place and $(n+1)^{th}$ place is about $\frac{1.44}n$ points in this system)
A: 
I wonder how multiplying the scores (let's call them x, y and z) leads to a different ranking than simply adding them.

I don't see how the inequality you are trying to prove relates to your question. In general multiplication can lead to completely different rankings than addition. The closer the numbers are the higher their product will be. For example 
A scores 2 / 2 / 3 and
B scores 1 / 1 / 7.
If we consider sums, then A has 7 points and B 9 points... but for the products
A has 12 points and B 7 points.
(a quick first approach here is noticing that a score of 1 has no contribution to the final score when you are multiplying).
A: The fact that the partial scores are being multiplied means that competitors will do better when they focus on certain aspects at the cost of others, rather than trying to be 'well-rounded'.
For example, if person A is really good at Lead (1 points), but fairly weak at both Speed (7 points) and Bouldering (7 points), for a total of $49$ points, then they will end up in a better position than the more 'well-rounded' person B who has 4 points on Lead, 4 on Speed and 4 on Bouldering, for a total of $64$ points. If we had added the scores, A would have had $15$ points, and B only $12$, and B would have ended up in a better position.
In short, the multiplication scoring system rewards 'specialization', whereas the addition scoring system would reward 'well-roundedness'.
I am wondering if the organizers did this intentionally, so as to try and get different kinds of climbers (one more speedy, another more technical, etc.), the variety of which may be more exciting for the audience to watch.
