Are there any integral solutions to this inequality?

Question

Are there any integral solutions to this inequality?

$$\frac{n\sqrt{3} + 1}{n\sqrt{3}} + {\left(\frac{2n}{n + 1}\right)}^{1/2} < 1 + \sqrt{3}$$

WolframAlpha appears to give an inconsistent result.

I will outline an algebraic approach in my answer.

Update: It turns out that my algebraic method (as well as the other answers) are strong enough to let us find solutions. I will post a follow-up question.

Answer 1

This is only a partial solution, but here is what I would do:

1. $1+\sqrt{3} \approx 2.732$, therefore, the solutions are bounded by this.
2. As $n \rightarrow \infty$, the left hand side goes towards: $\frac{n\sqrt{3}+1}{n\sqrt{3}}+\left(\frac{2n}{n+1}\right)^{1/2} \rightarrow \frac{n\sqrt{3}}{n\sqrt{3}}+\left(\frac{2n}{n}\right)^{1/2} \rightarrow 1+\sqrt{2} \approx. 2.414$ This is less than the right hand side, so some integer should exist, given a large enough $n$.
3. As a computer geek, I would just do some kind of programmatic search for an answer, starting with really big guess (say 1 billion), and doubling it every time until the answer is true.

Because I assume you are looking for a more mathematical approach, here is my next set of steps. $$\frac{n\sqrt{3}+1}{n\sqrt{3}}+\left(\frac{2n}{n+1}\right)^{1/2} < 1+\sqrt{3}$$ $$1+\frac{1}{n\sqrt{3}}+\sqrt{2}\sqrt{\frac{n}{n+1}} < 1+\sqrt{3}$$ $$n\sqrt{3}*\sqrt{n+1}+1*\sqrt{n+1}+\sqrt{2}\sqrt{n}*n\sqrt{3} < (1+\sqrt{3})*n\sqrt{3}*\sqrt{n+1}$$ We now have all the denominator terms removed, so we can simplify, if possible, to find the answer.

Answer 2

Simplify: $\sqrt{3}-\dfrac{1}{n\sqrt{3}} > \sqrt{\dfrac{2n}{n+1}}\iff 3 - \dfrac{2}{n}+\dfrac{1}{3n^2}> \dfrac{2n}{n+1}$. Without further checking using limit ($3 > 2$), the $LHS$ > $RHS$ starting at some $n$. Thus it is true.

Answer 3

$$\frac{n\sqrt{3} + 1}{n\sqrt{3}} + {\left(\frac{2n}{n + 1}\right)}^{1/2} < 1 + \sqrt{3}$$ $${\left(\frac{2n}{n + 1}\right)}^{1/2} < 1 + \sqrt{3} - \left(\frac{n\sqrt{3} + 1}{n\sqrt{3}}\right) = \frac{n\sqrt{3} + 3n - n\sqrt{3} - 1}{n\sqrt{3}} = \frac{3n - 1}{n\sqrt{3}}$$

Squaring both sides, I get

$$\frac{2n}{n + 1} < {\left(\frac{3n - 1}{n\sqrt{3}}\right)}^2 = \frac{9n^2 - 6n + 1}{3n^2}.$$

Cross-multiplying, we obtain

$$6n^3 < 9n^3 + 3n^2 - 5n + 1.$$

The final resulting inequality is

$$0 < 3n^3 + 3n^2 - 5n + 1$$ which is satisfied by all integers $n \geq 1$.