This is a problem from an Ukrainian math competition which was held on 28 January 2024.
Find the smallest real number $M$ for which
$$\{a\} + \{b\} + \{c\} ≤ M$$
for any positive real numbers $a, b, c$ such that
$$abc=2024$$
where $\{x\}$ denotes the fractional part of $x$, for example: $$\{3.14\} = 0.14$$
I'm assuming the inequality of harmonic, geometric, arithmetic and quadratic means could be used to solve this, but have no idea how to actually apply it. Having all the means
$$x_{harmon}≤x_{geom}≤x_{arithm}≤x_{quadr}$$
which, for positive real numbers a, b and c is
$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}≤\sqrt[3]{abc}≤\frac{a + b + c}{3}≤\sqrt{\frac{a^2+b^2+c^2}{3}}$$
we can get (from the last inequality) to
$$\{a\} + \{b\} + \{c\} ≤ \sqrt{3(a^2+b^2+c^2)} – ([a]+[b]+[c])$$
where $[a] = a – \{a\}$ is a whole part of a number. We now have sum of those three fractions separated on the left, but I'm not sure what can be done further from this point.
There is an suggested answer for this problem. It's $2 + (2024/2025)$ which is approximately $2.99950….$ But what is the logic behind it and is it even correct – I'm not sure.
Best Answer
As D S observes, one can see that $M \ge 2 + \frac{2024}{2025}$ by taking any integers $n, m$ with $nm = 2025$ and looking at the limit of $(a, b, c) = (n - \epsilon, m - \epsilon, \frac{2024}{(n - \epsilon)(m - \epsilon)})$ as $\epsilon \to 0$.
This example, along with my comment about the fact that near-optimal triples tend to be "a bit less than a triple of integers multiplying to an integer close to $2024$", already sort of suggests why this might be optimal. The idea is that among triples $(a, b, c)$ with $\lceil a \rceil = n$, $\lceil b \rceil = m$ and $\lceil c \rceil = k$, the best you can do is $\{a\} + \{b\} + \{c\} \le 2 + \frac{2024}{mnk}$, with an example like in the first paragraph. Once we've shown that, it's clear that the largest $M$ we can hope for is given by making $mnk$ as small as possible, and hence taking $mnk$ to be $2025$.
To prove this fact, we exactly need to prove the following:
Lemma. If $A, B, C \ge 0$ and $x, y, z \in [0, 1]$, then \begin{equation*} x + y + z \le 2 + \frac{(A + x)(B + y)(C + z)}{(A + 1)(B + 1)(C + 1)} \end{equation*}
Proof. Fix $A, B, C$, and consider the function \begin{equation*} f(x, y, z) = -(x + y + z) + 2 + \frac{(A + x)(B + y)(C + z)}{(A + 1)(B + 1)(C + 1)} \end{equation*} We would like to show that $f(x, y, z) \ge 0$ for $x, y, z \in [0, 1]$. We'll do this by minimising $f$. Observe that $f$ is actually decreasing in each of its arguments. (eg if $x_1 \le x_2$, then $f(x_1, y, z) \ge f(x_2, y, z)$, and so on - the point being that with two arguments fixed, $f$ is a linear function with non-positive gradient). So $f$ is minimised when $(x, y, z) = (1, 1, 1)$. But $f(1, 1, 1) = 0$. Hence $f(x, y, z) \ge 0$, and we are done.
(To apply this lemma, take $A = \lfloor a \rfloor, B = \lfloor b \rfloor, C = \lfloor c \rfloor$ and $x = \{a\}, y = \{b\}, z = \{c\}$).
I have been a little bit loose with floors and ceilings here, but that is fine, because if any of $a, b, c$ is an integer then of course $\{a\} + \{b\} + \{c\} \le 2$, so that case isn't even worth thinking about.