Solving inequality involving square root and division by logarithm: $\sqrt n<\frac{n}{\log(n)}-2$ I would like to solve the inequality $\sqrt n<\frac{n}{\log(n)}-2$. for some reason I had never done this before. This is clearly the same  as $\frac{n}{\log(n)}-\frac{n}{\sqrt{n}}>2$. Which is the same as $\frac{\sqrt n)(\sqrt n-\log(n))}{\log(n)}>2$. But here I get stuck. Thanks in advance.
 A: It doesn’t look like it has a closed-form solution, other than in terms of the two solutions of $\frac{\sqrt n)(\sqrt n-\log(n))}{\log(n)}{\color{blue}=}2$.
Solving numerically with Mathematica, the inequality holds for $0<n<1.6543858...$ and for $n>18.138863...$. 

The calculations here show that Mathematica finds no closed form for the two intersection points and denotes them simply as the points where $\frac{\sqrt n)(\sqrt n-\log(n))}{\log(n)}=2$
I don’t know what numerical technique Mathematica used, but the function is well behaved near its two solutions, so no fancy method beyond guessing and checking is needed. For the larger solution, for example, you could try $n= 18$, based on the graph, see that the curve lies below $y=2$, then try $y=18.5$, see that it lies above, then $y=18.2$, and so on. 

A: If you want to solve the inequality with minimal use of computational assistance it is going to be messy.  Here is one thing you could do.
Note that when $t$ is positive we have
$$\log t\le t-1\ .$$
Now
$$\log n=4\log(n^{1/4})$$
and so
$$\eqalign{(\sqrt n+2)\log n
  &\le4(n^{1/2}+2)(n^{1/4}-1)\cr
  &<4n^{3/4}-4n^{1/4}(n^{1/4}-2)\ .\cr}$$
Now if $n^{1/4}>4$ then $4n^{3/4}<n$ and $n^{1/4}-2>0$, so this last expression is less than $n$.
Therefore the inequality is true when $n>256$, and if $n$ is an integer variable you can check separately for $n=1,2,\ldots,256$.
I told you it would be messy ;-)
