Can you prove that $[(x^2 + 2x + 1)\log(x) – x^2\log(x+1)]/2\log(x)\log(x+1)$ > 1 for all x > 2? Can you prove that $\frac{(x^2 + 2x + 1)\log(x) – x^2\log(x+1)}{2\log(x)\log(x+1)} &gt; 1$ for all $x &gt; 2$?
The target inequality can be rewritten as
\begin{align*}
\frac{(x+1)^2}{2\log(x+1)} &gt; \frac{x^2}{2\log(x)} + 1 \iff f(x+1) &gt; f(x) + 1
\end{align*}
where $f(x) = \frac{x^2}{2\log(x)}$. By Taylor series,
\begin{align*}
f(x+1) = f(x) + \color{blue}{f'(x) + \frac{1}{2}f''(\xi)}
\end{align*}
for some $\xi \in (1, x)$. We strive to prove the blue portion is $&gt; 1$. Note that
\begin{align*}
f'(x) = \frac{x}{\log(x)} - \frac{x}{2\log^2(x)}, \qquad f''(x) = \frac{2(\log(x)-0.75)^2+0.875}{2 \log^3(x)}
\end{align*}
Note that $f''(x) &gt; 0$ for all $x &gt; 2$ (in fact, $x &gt; 1$), so therefore $f'(x)$ is increasing, and
\begin{align*}
f'(x) &gt; f'(2) = \frac{\log(4)-1}{\log^2(2)} &gt; 0.80 \quad \text{for all } x &gt; 2
\end{align*}
Not strong enough, but note that $f'(e) = e/2 = 1.359 &gt; 1$, so if we can prove $\frac{1}{2}f''(\xi) &gt; 0.2$ for $\xi \in [2, e]$, then we are done. Indeed,
\begin{align*}
\frac{1}{2}f''(\xi) \ge \frac{0.875}{4 \log^3(\xi)} \ge \frac{0.875}{4 \log^3(e)} = 0.21875 &gt; 0.2
\end{align*}
Here is my answer, people can correct me if I am wrong.

$$\frac{[(x+1)^2\log{(x)}-x^2\log{(x+1)}]}{2\log{(x)}\log{(x+1)}}$$

$$\frac{[\log(x^{{(x+1)}^{2}})-\log{{({{(x+1)}^x}^2}}{}]}{\log(x^2)\log(x+1)}$$

$$\frac{\log(x^{{(x+1)}^{2}})}{\log(x^2)\log(x+1)} \frac{\log(x^2)\log(x+1)}{\log{{(x+1)^x}^2}}$$

$$\frac{{(x+1)}^{2}\log(x)}{x^2\log(x+1)}$$

I think that the last form is always bigger than 1 if x > 2, just plug-in 2 or any greater number.