<s>[INST] Upper Bound for $\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$ I was reading a solution for an analysis problem and they argued that $$\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$$ for $x\geq 1$ and $n\geq 1$,  $x$ a real number and $n$ a natural number. Why is that?
 [/INST] We have $\frac{\ln(1+n x)}{1+x^2 \ln n} \leq \frac{1+\ln(x)+\ln(2)}{x^2}$ such that $n \in \mathbb{N} \geq 1$ and $x \in \mathbb{R} \geq 1$,
Using the fact that $\ln x+\ln y = \ln(x y)$ we get that $\frac{\ln(1+n x)}{1+x^2 \ln n} \leq \frac{1+\ln(2x)}{x^2}$.
We will assume that $n\geq 3$ so $1+x^2 \ln n$ and $x^2$ are positive numbers so we multiply by them does not change the inequality.
We arrive at $x^2 \ln(1+ n x) \leq (1+x^2 \ln n)(1+\ln (2x))$
Which is just $x^2 \ln (1+n x) \leq 1+x^2 \ln n+\ln(2x) + x^2 \ln(2x) \ln n$
Now since $1+\ln(2x)$ is positive for all $x\geq 1$ we can treat them as $0$ to strengthen the inequality and get that $x^2 \ln (1+n x) \leq x^2 \ln n + x^2 \ln(2x) \ln n$ Now divide by $x^2 \ln n$
We get that $\frac{\ln(1+n x)}{\ln n} \leq 1+\ln(2x)$ which is just $\frac{\ln(1+n x)-\ln n+\ln n}{\ln n} \leq 1+\ln(2x)$
Using the fact that $\ln x-\ln y = \ln(x/y)$ we get that $\frac{\ln(\frac{1+n x}{n})}{\ln n }+1 \leq 1+\ln(2x)$
Which is just $\frac{\ln(\frac{1}{n}+x)}{\ln n}\leq \ln(2x)$
Since $n\geq 3$ its easy to see that $\ln n \geq 1$ so $\ln(\frac{1}{n}+x) \geq \frac{\ln(\frac{1}{n}+x)}{\ln n}$
So to strengthen the inequality we get that $\ln(\frac{1}{n}+x) \leq \ln(2x)$
And since $\ln$ function is increasing then the inequality is true whenever $\frac{1}{n}+x \leq 2x$ which is just $\frac{1}{n} \leq x$ and since $x\geq 1$ and $n\geq 3$ so $\frac{1}{n} \leq \frac{1}{3}$ the we get that $\frac{1}{n} \leq \frac{1}{3} \leq 1 \leq x$ which is always true.
Now we are left with cases $n=1,2$.
For $n=1$ we get that $\ln(1+x) \leq \frac{1+\ln(2x)}{x^2}$ which is obviously false for all $x\geq 2$
For $n=2$ we get that $\frac{\ln(1+x)}{1+x^2 \ln 2} \leq \frac{1+\ln(2x)}{x^2}$
Which is also false for all number $x\geq6$ because the terms the play the major rule are $x^2$ and $x^2 \ln 2 \approx 0.693 x^2$ and since $x^2$ is much bigger than $0.693 x^2$ so the inequality obviously  will be false from some point on.
thus concluding that the proof is true whenever $n\geq3$ and $x\geq1$.
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