# How prove this $\left[\frac{n}{\sqrt{3}}\right] +1\ge \frac{n^2}{\sqrt{3n^2-\frac{50}{9}}}$

For any integer $n\ge 2$ ,Prove that: $\left[\frac{n}{\sqrt{3}}\right] +1\ge \frac{n^2}{\sqrt{3n^2-k}}$ .

when $k=\dfrac{50}{9}$

This $k$ is folowing idea found it.

let $n=2$,then we have $$2\ge \dfrac{4}{\sqrt{12-k}}\Longrightarrow k\le 8$$

let $n=4$ we have $$2+1\ge\dfrac{16}{\sqrt{48-k}}\Longrightarrow k\le\dfrac{176}{9}$$ $n=5$,then we have $$2+1\ge \dfrac{25}{\sqrt{27-k}}\Longrightarrow k\le\dfrac{50}{9}$$ let $n=6$,we have $k\le 27$

let $n=10$,then we have $$6\ge\dfrac{100}{\sqrt{300-k}}\Longrightarrow k\le\dfrac{200}{9}$$

so I guess the maximum $k=\dfrac{50}{9}$, I want use @achille hui idea to prove it,and I found his methods can't use for this stronger inequality $$\dfrac{n^2}{\sqrt{3n^2-\dfrac{50}{9}}}\ge\dfrac{n^2}{\sqrt{3n^2-5}}$$simaler problem

• Sure you can. Since $\frac{n^2}{3q_n^2} \to 1$ as $n \to \infty$. You know there is a $N$ such that $2 \frac{n^2}{q_n^2} \ge \frac{50}{9}$ for all $n \ge N$. It is not hard to check $N = 45$ works. What remains is to verify the newer inequality works for $2 \le n \le 44$ by brute force. – achille hui Nov 14 '14 at 12:28
• And here's the brute force m.wolframalpha.com/input/… – Macavity Nov 14 '14 at 16:34