2
$\begingroup$

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 $x$

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

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 $5$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.