Show that there exists a positive constant $C$ such that, for all positive reals $a$ and $b$ with $a + b$ being an integer, we have $$\left\{a^3\right\} + \left\{b^3\right\} + \frac{C}{(a+b)^6} \le 2. $$Here $\{x\} = x - \lfloor x\rfloor$ is the fractional part of $x$. Proposed by Li4 and Untro368.
Problem
Source: 2023 Taiwan TST Round 3 Independent Study 1-A
Tags: integer and fractional part, algebra
sqing
26.04.2023 11:37
Find out the largest constant $C \in \mathbb{R}^+$, so that if $a+b \in \mathbb{N}^+$, where $a,b \in \mathbb{R}^+$, we have $$\left \{ a^2 \right \}+\left \{ b^2 \right \} \le 2 - \frac{C}{\left ( a+b \right )^2}.$$Where $\left \{ . \right \}$ denotes the fractional part. 2018 Peking University Summer Camp https://artofproblemsolving.com/community/c6h1885856p12848213 https://artofproblemsolving.com/community/c6h1696769p10880563 https://artofproblemsolving.com/community/c6h2407965p19740400
CANBANKAN
22.05.2023 08:27
Nice problem! Let $N=a+b \in \mathbb{N}^+$.
Let $\beta = X^{1/3} > N/2$ where $X\in \mathbb{N}$. Let $C=10^{-4}$ Then for any integer $k$,
$$\left| \beta^2 - N\beta - \frac{k}{3N}\right| \ge CN^{-7}$$
Let $\alpha = \beta^2 - N\beta$. Then check that $$\alpha^3 + 3XN\alpha + X(N^3-X) = 0$$
Let $f(x) = x^3 + 3XNx + X(N^3-X)$. Then $f'(x) = 3x^2 + 3XN$.
Assume for contradiction that there exists $k$ such that $\left| \beta^2 - N\beta - \frac{k}{3N}\right| <CN^{-7}$. By intermediate value theorem, there exists $c$ between $k/3N$ and $\alpha$ such that
$$ |f(k/3N) - f(\alpha)| = |k/3N - \alpha| \cdot | f'(c)| \le |k/3N - \alpha| \cdot \max_{c\in [\alpha-CN^{-7}, \alpha+CN^{-7}]} | f'(c)|$$
Note that the max exists because $|f'|$ is continuous and $[\alpha-N^{-7}, \alpha+N^{-7}]$ is compact. It's easy to see that $|f'| \le 50N^4$ in this interval. Therefore
$$ \frac{1}{(3N)^3} \le |f(k/3N) - f(\alpha)| \le |k/3N - \alpha| (50N^4)$$
This implies that $|\frac{k}{3N}-\alpha| \ge 10^{-4}N^{-7}$, as desired.
I claim $C= 10^{-8}$ works.
WLOG $0<a<b$. Let $b^3 = X - \epsilon$, where $\epsilon < 10^{-2}N^{-6}$. Hence
$$ a^3 - (N-\beta)^3 = (N-b)^3 - (N-\beta)^3 = (\beta-b)( (N-b)^2 + (N-b)(N-\beta) + (N-\beta)^2) \le (\beta-b) (b^2+b\beta+\beta^2) = \beta^3-b^3 = \epsilon$$.
Therefore, $\{a^3\} \le \{ (N-\beta)^3\} + \epsilon$. By the above lemma, we can show that $\{ (N-\beta)^3\} \le 1- |3N (\beta^2-N\beta) - k| = 1-3N | (\beta^2-N\beta) - k/(3N)| \le 1-10^{-4}N^{-6}$. Therefore, $$\{a^3\} + \{b^3\} < 2 - (10^{-2} + 10^{-4}) N^{-6}$$
Tintarn
31.05.2023 12:28
In general, one can show in a similar way that for any $k$ there exists a constant $C_k$ such that whenever $a,b>0$ with $a+b$ an integer, then \[\{a^k\}+\{b^k\}+\frac{C}{(a+b)^{k^2-k}} \le 2.\]