Research question - prove inequality for $n \log n$. The following inequality has come up in some research and appears true when plotted, but I'm not sure how to to prove it. If $g(z) = z \log z$, then for $x>3$ ($x$ real),
$$
g(x) < g(x-1)\cdot \left(1 + \frac{g(x-1) - g(x-2)}{g(x-1)-1} \right).
$$
 A: I believe the following outline works:

*

*It suffices to show that the function $\dfrac{g(x-1)-g(x-2)}{g(x)-g(x-1)} g(x-1)-g(x-1)+1$ is positive for $x>3$ (since then one can multiply through by $\dfrac{g(x)-g(x-1)}{g(x-1)-1}$ which is positive).

*The generalized mean value theorem tells us that if $f$ and $g$ are differentiable, then
$$
\frac{f(x)-f(x-1)}{g(x)-g(x-1)} = \frac{f'(\xi)}{g'(\xi)}
$$
for some $\xi\in(x-1,x)$. Applying this with $f(x) = g(x-1)$ gives
$$
\frac{g(x-1)-g(x-2)}{g(x)-g(x-1)} = \frac{g'(\xi-1)}{g'(\xi)} = \frac{1+\log(\xi-1)}{1+\log\xi}
$$
for some $\xi\in(x-1,x)$. And since $\dfrac{1+\log(\xi-1)}{1+\log\xi}$ is increasing, we can replace it by its lower bound $\dfrac{1+\log(x-2)}{1+\log(x-1)}$.

*We have thus shown that it suffices to show that $\dfrac{1+\log(x-2)}{1+\log(x-1)} g(x-1)-g(x-1)+1$ is positive for $x>3$, or equivalently that $1 + \log(x-1) - g(x-1) \big(\!\log(x-1) - \log(x-2)\big)$ is positive there (since we can then divide by $1+\log(x-1)$ which is positive).

*By the ordinary mean value theorem, $\log(x-1) - \log(x-2) = \dfrac1{\eta}$ for some $\eta\in(x-2,x-1)$, and in particular $\log(x-1) - \log(x-2) < \dfrac1{x-2}$.

*It therefore suffices to show that $1 + \log(x-1) - \dfrac{g(x-1)}{x-2} > 0$ is positive for $x>3$, or equivalently that $x - 2 - \log(x-1)$ is positive there (since we can then divide by $x-2$ which is positive).

*But finally this is a simple calculus calculation, since $x-2-\log(x-1)=0$ when $x=2$ and $\dfrac d{dx}(x-2-\log(x-1)) = 1-\dfrac1{x-1}>0$ for $x>2$.

A: Alternative solution:
Fact 1: $\frac{2u}{2+u} \le \ln (1 + u) \le \frac{2u + u^2}{2 + 2u}$ for all $u \ge 0$.
(The proof is easy. Hint: Take derivative.)
Now, first, by Fact 1, we have
$$g(x) - x\ln (x-1) = x \ln(1 + \tfrac{1}{x-1})
\le x \cdot \frac{2x-1}{2x(x-1)} = \frac{2x-1}{2(x-1)}$$
which results in
$$g(x) \le x\ln(x-1) + \frac{2x-1}{2(x-1)}.$$
Second, by Fact 1, we have
$$(x-2)\ln(x-1) - g(x-2) = (x-2)\ln(1 + \tfrac{1}{x-2})
\ge (x-2) \cdot \frac{2}{2x-3} = \frac{2(x-2)}{2x-3}$$
which results in
$$g(x - 2) \le (x-2)\ln(x-1) - \frac{2(x-2)}{2x-3}.$$
Third, it suffices to prove that (note: $g(x-1) - 1 > 0$)
$$x\ln(x-1) + \frac{2x-1}{2(x-1)} 
< g(x-1) \cdot \left(1 + \frac{g(x-1) - [(x-2)\ln(x-1) - \frac{2(x-2)}{2x-3}]}{g(x-1)-1} \right)$$
that is
$$\frac{1}{2(2x-3)[(x-1)\ln(x-1)-1]}\left(4(x-1) - \frac{1}{x-1} - \ln(x-1)\right) > 0$$
which is true since $x-1 \ge \ln(x-1)$ and $x - 1 \ge \frac{1}{x-1}$.
We are done.
