Comparing two close numbers

Let $f(x)=\frac{\sqrt{x+1}\ln(1+x)}{x},$ where $x>0$.

Thus, $$f'(x)=\frac{\left(\frac{\ln(1+x)}{2\sqrt{1+x}}+\frac{1}{\sqrt{1+x}}\right)x-\sqrt{1+x}\ln(1+x)}{x^2}=\frac{2x-(x+2)\ln(1+x)}{2x^2\sqrt{1+x}}\leq0$$ because $$\left(\ln(1+x)-\frac{2x}{x+2}\right)'=\frac{x^2}{(x+1)(x+2)^2}\geq0.$$ Id est, $f$ decreases and for all $n>0$ we obtain: $$f\left(\frac{1}{n+1}\right)>f\left(\frac{1}{n}\right)$$ or $$\frac{\sqrt{\frac{1}{n+1}+1}\ln\left(1+\frac{1}{n+1}\right)}{\frac{1}{n+1}}>\frac{\sqrt{\frac{1}{n}+1}\ln\left(1+\frac{1}{n}\right)}{\frac{1}{n}}$$ or $$\sqrt{(n+1)(n+2)}\cdot\ln\frac{n+2}{n+1}>\sqrt{n(n+1)}\cdot\ln\frac{n+1}{n}$$ or $$\left(\frac{n+2}{n+1}\right)^{\sqrt{n+2}}>\left(\frac{n+1}{n}\right)^{\sqrt{n}}.$$

Now, take $n=10.$