Upper Bound for $\frac{\ln(1+nx)}{1+x^2\ln(n)}\leq \frac{1+\ln(x)+\ln(2)}{x^2}$

Question

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?

Answer 1

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 $x\geq1$.

Answer 2

First the identity definitely doesn't hold for $n=1$ so let's assume $n\geq 2$, then we have $$ \begin{align*} \ln(1 + nx) &= \ln(nx) +\ln(1+1/nx) \\ &\leq \ln(nx) + 1/nx =\frac{ x^2 \ln (nx) + x/n }{x^2}. \end{align*} $$ Now \begin{align*} x^2\ln(nx) + x/n &= \left[ 1+x^2\ln(n) \right] \left( \frac{\ln(n)}{\ln(n) + x^{-2} } + \frac{\ln(x) }{\ln(n) + x^{-2} } + \frac{1}{nx(\ln(n) + x^{-2} )} \right) \end{align*} giving \begin{align*} \frac{ \ln( 1 + nx) }{ 1 + x^2 \ln(n) } &\leq \frac{1}{x^2}\left( \frac{x^2 \ln(n) }{x^2 \ln(n) + 1 } + \frac{x^2 \ln(x) }{x^2 \ln(n) + 1 } +\frac{x^2 }{nx ( x^2 \ln(n) + 1 )}\right) \\ &= \frac{\ln(n) }{x^2 \ln(n) + 1 } + \frac{\ln(x) }{x^2 \ln(n) + 1 } +\frac{1}{nx ( x^2 \ln(n) + 1 )} \\ &\leq \frac{1}{x^2} \left( 1 + \frac{\ln(x)}{\ln(n)} + \frac{1}{n \ln (n) }\right) \end{align*} For $n \geq 3$ this last bound is actually tighter than the one you are trying to prove, and in fact your original inequality fails for $n=2$ but I don't have an immediate assertion of that fact. Will come back to it.

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Just to finish the answer off for the case $n=2$ the original inequality is equivalent to $$ \frac{ \ln(1 + 2x) }{ 1 + \ln(2 x) } \leq \frac{1 + x^2 \ln(2) }{x^2 } = \frac{1}{x^2} + \ln(2), \qquad x \geq 1. $$ which is true for $x = 1$, but $$ \lim_{x \rightarrow \infty}\frac{\ln(1 + 2x)}{1 + \ln(2x) } = 1, \qquad \lim_{x \rightarrow \infty} \frac{1}{x^2} + \ln(2) =\ln(2) < 1, $$ and so the claim is false for $n=1, 2$.