# How prove this inequality $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{n^2}{\sqrt{3n^2-5}}$

let $k$ is postive integer,and for any postive integer $n\ge 2$,

show that: $$\left[\dfrac{n}{\sqrt{3}}\right]+1>\dfrac{n^2}{\sqrt{3n^2-5}}>\dfrac{n}{\sqrt{3}}$$

where $[x]$ is the largest integer not greater than $x$

• Using $n=5$, you get $k \le 5$. So we just need to prove $k=5$ works. – Macavity Nov 14 '14 at 5:23
• $\Big\lfloor \dfrac{n}{\sqrt3} \Big\rfloor+1 > \dfrac{n^2}{\sqrt{3n^2-5}} > \dfrac{n}{\sqrt3}$ holds true, though the left inequality seems tough to prove. – Macavity Nov 14 '14 at 9:42

Let $q_n = \left\lfloor \frac{n}{\sqrt{3}}\right\rfloor + 1$. When $n \ge 11\sqrt{3}$, we have

$$\frac{n}{\sqrt{3}q_n} \ge \frac{\frac{n}{\sqrt{3}}}{\frac{n}{\sqrt{3}}+1} = 1 - \frac{\sqrt{3}}{n+\sqrt{3}} \ge \frac{11}{12} \quad\implies\quad \frac{n^2}{3q_n^2} \ge \left(\frac{11}{12}\right)^2 > \frac{5}{6}$$ Be definition, $\left\lfloor \frac{n}{\sqrt{3}}\right\rfloor$ is the largest integer less than or equal to $\frac{n}{\sqrt{3}}$. This implies

$$\frac{n}{\sqrt{3}} < q_n \quad\implies\quad 3q_n^2 - n^2 > 0 \quad\implies\quad 3 q_n^2 - n^2 \ge 2$$

The last inequality is true because $3q_n^2 - n^2$ is an integer and there is no integer solution for the equation $3 q^2 - n^2 = 1$.

Combine these, we find for any $n \ge 20 > 11\sqrt{3}$, we have

$$3 n^2 - \frac{n^4}{q_n^2} = 3\left(\frac{n^2}{3q_n^2}\right)(3q_n^2 - n^2) > 3 \left( \frac{5}{6}\right) 2 = 5$$ This leads to $$3n^2 - 5 > \frac{n^4}{q_n^2} \quad\iff\quad q_n > \frac{n^2}{\sqrt{3n^2 -5}} \quad\text{ for } n \ge 20 \tag{*1}$$

By brute force, one can verify RHS$(*1)$ also work for $2 \le n \le 19$. As pointed out by Macavity in comment, the largest admissible $k$ for $n = 5$ is $5$. This means the maximum value of $k$ which works for all $n$ is indeed $5$.

• very very nice!thank you,+1 – math110 Nov 14 '14 at 10:21
• Nice... $+ 1$ – Macavity Nov 14 '14 at 12:41