Proof that $\ln(n^2)(\ln(n) - 1) < n$ for all $n\in\mathbb{N}$ I would like to know which proof strategy to use when proving the next inequality:
$\ln(n^2)(\ln(n) - 1) < n,\quad\forall n \in \mathbb{N}$. I have been trying to use this two proved inequalities $\dfrac{n}{n+1} < \ln(n+1) < n$ ,but it did not give me solution. 
 A: Define a function $u$ on $x>0$ by $u(x)=x-\log(x^2)(\log(x)-1)=x-2\log(x)^2+2\log(x)$. The goal is to prove that $u(n)>0$ for every positive integer $n$, let us prove the stronger statement that $u(x)\ge1$ for every real number $x\ge1$. 
To do so, first note that $u'(x)=v(x)/x$ with $v(x)=x-4\log(x)+2$ and that $v'(x)=1-4/x$, hence $v$ is decreasing on $(1,4)$ and increasing on $(4,+\infty)$. 
Since $v(4)=6-8\log(2)=.4548>0$, $v>0$ everywhere and $u$ is increasing. In particular, $u(x)\ge u(1)=1$ for every $x\ge1$. 
Finally, for every positive integer $n$,
$$
\log(n^2)(\log(n)-1)\le n-1<n.
$$
A: Make the substituting $x = \log n$, so $x \geq 0$. The inequality now reads
$$ 2x(x-1) < \exp(x). $$
All we have to do know is to take the Taylor expansion of $\exp(x)$ and stop at the right place. For example,
$$ 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} < \exp(x), $$
and it so happens that when $x \geq 0$,
$$ 2x(x-1) < 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}. $$
This can be verified formally using e.g. Sturm sequences (or by finding explicitly all the roots using the quartic formula).
A: Here's a simple approach. Verify the inequality for the base cases $n=1$ and $n=2$; we will assume $n \geq 3$ from now on. Let us now make the substitution $x = \ln n$ (where $x \geq \ln 3 > 1$), and rewrite the inequality as $e^x \geq 2x(x-1)$. 
Using the famous inequality $e^{t} \geq t+1$, we get 
$$
e^{x- \frac{3}{2}} \geq x- \frac{1}{2}.
$$
for all real $x$. Integrating between the limits $0$ and $x > 0$, we get:
$$
e^{x- \frac{3}{2}} - e^{-\frac{3}{2}} \geq \frac{1}{2}(x^2-x).
$$
Rearranging this slightly (and dropping one term), we get 
$$
e^x \geq \frac{e^{3/2}}{4} \cdot 2x(x-1),
$$
which implies the claim since $\frac{e^{3/2}}{4} \geq 1.1 > 1$. (Note that the final step is valid only if $2x(x-1)$ is positive, but this is the true since $x \geq 1$.) 
