Prove upper bound to $\lfloor\log_k{x} \rfloor$

Question

I have the following inequality conjectured, but I'm having quite a tough time proving it... Any ideias? $$\lfloor \log_k{x}\rfloor \leq 1+\frac{x^2}{k}\;\;\;\forall x \in \mathbb{Z}^+$$

Answer 1

You're conjecture is not true for small values of $k$. For example $k=1.01$ and $1.02 < x < 16$, the inequality is false. See this graph: https://www.desmos.com/calculator/ggiaiqxenc. Where ever the graph is negative the inequality has failed.

EDIT

Removing the floor function, the inequality becomes true if $k \geq 1.130936$ (approximately)

This can be calculated using the technique in another answer. Namely find the minimum of $1+\frac{x^2}{k}-\ln_kx$. The derivative is $\frac{2x}{k}-\frac1{x\ln k }$. So the minimum occurs when $\frac{2x}{k}=\frac1{x\ln k } \Longrightarrow x=\sqrt\frac{k}{2\ln k}$. Substituting this back in gives: $1+\frac{k^2}{2\ln k} - \frac{\ln\sqrt\frac{k}{2\ln k}}{\ln k}=0$. Solving this numerically gives $\approx 1.130936$

Answer 2

Recall that $$ \lfloor log_k{x}\rfloor \leq \log_k(x) $$ Show that the minimum of the function $$ f(x) :=x^2/k + 1 - \log_k(x) $$ on $(0,\infty)$ is >0. This is usually solved by calculus. Then $\log_k(x)\leq 1+x^2/k$.