Taiwan IMO Selection Camp, 2023

We show that the claim holds when $C = \frac{1}{54}$:

Let $n = a + b$, and without loss of generality assume $a \ge b$. Note that the function $f(x) = x^3$ is an increasing convex function, so if $a_0 + b_0 = n$ and $a_0 - b_0 \ge a - b$, we have
$$a^3 + b^3 \le a_0^3 + b_0^3. \quad (1)$$
In particular, when $\lfloor a^3 \rfloor = \lfloor a_0^3 \rfloor$ and $\lfloor b^3 \rfloor = \lfloor b_0^3 \rfloor$,
$$\{a^3\} + \{b^3\} \le \{a_0^3\} + \{b_0^3\}.$$
So it suffices to consider the two cases $\{a^3\} \to 1^-$ or $\{b^3\} = 0$. The latter is obvious:
$$\{a^3\} + \{b^3\} < 1 < 2 - \frac{C}{n^6}.$$
For the former, suppose $(a + \varepsilon)^3 = k < n^3$ is a positive integer, in which case $b = n - \sqrt[3]{k} + \varepsilon$. Since $\varepsilon$ is sufficiently small, by (1),
$$\begin{aligned} \{a^3\} + \{b^3\} &= a^3 + b^3 - \lfloor a^3 \rfloor - \lfloor b^3 \rfloor \\ &\le k + (n - \sqrt[3]{k})^3 - (k - 1) - \lfloor (n - \sqrt[3]{k})^3 \rfloor \\ &= 1 + \{(n - \sqrt[3]{k})^3\}. \end{aligned}$$
So it suffices to prove
$$1 + \{(n - \sqrt[3]{k})^3\} \le 2 - \frac{C}{n^6}.$$
If $(n - \sqrt[3]{k})^3$ is an integer, the claim is obvious. If not, the above inequality is equivalent to
$$\frac{C}{n^6} \le 1 - \{(n - \sqrt[3]{k})^3\} = \{-(n - \sqrt[3]{k})^3\} = 3n^2\sqrt[3]{k} - 3n\sqrt[3]{k^2} - m,$$
where $m = \lfloor 3n^2\sqrt[3]{k} - 3n\sqrt[3]{k^2} \rfloor \ge 0$. Consider the following identity
$$(x + y + z)(x^2 + y^2 + z^2 - yz - zx - xy) = x^3 + y^3 + z^3 - 3xyz,$$
substituting $x = 3n^2\sqrt[3]{k}$, $y = -3n\sqrt[3]{k^2}$, $z = -m$ gives
$$0 < (1 - \{b^3\})(x^2 + y^2 + z^2 - yz - zx - xy) = 27n^3k(n^3 - k - m) - m^3.$$
Since the right-hand side is an integer, the left-hand side is at least 1. Therefore, from $\sqrt[3]{k} < n$ and $0 \le m \le 3n^2\sqrt[3]{k} < 3n^3$,
$$\begin{aligned} 1 - \{b^3\} &\ge \frac{1}{x^2 + y^2 + z^2 - yz - zx - xy} \ge \frac{1}{2(x^2 + y^2 + z^2)} \\ &\ge \frac{1}{2(9n^6 + 9n^6 + 9n^6)} = \frac{1}{54n^6}, \end{aligned}$$
and the proof is complete.