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We have the following inequality for the logarithm: given any $0<a<1$, there exists a constant $C_a$ such that $$\log (1+x) \leq C_a \frac{x}{(1+x)^a} $$ holds for all $x>0$. In other words, for each $a\in (0,1)$ we can find a uniform constant $C_a>0$ such that the above inequality holds for all $x>0$.

We can generalize this as $$ \frac{\log x - \log y}{x-y} \le C_a \frac{1}{y^{1-a}x^a}, $$ where $x>y$ and $0<a<1$.

I am trying to prove this inequality using the following integration method: $$ \frac{\log x - \log y}{x-y} = \int_{0}^{1} \frac{1}{(1-s)y+sx}\, \mathrm{d}s. $$ So I need to show that $$ \int_{0}^{1} \frac{1}{(1-s)y+sx}\, \mathrm{d}s \le C_a \frac{1}{y^{1-a}x^a}. $$

This looks like an interpolation inequality or log-convex type property. But I couldn't figure it out. So, my question is how to prove this using the above integral representation?

Any comments or hints would be helpful (or any reference).

And, what kind of functions satisfy this type of property (integral of $f$ can be bounded by some constant times $f(0)^{1-a}f(1)^{a}$)?

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    $\begingroup$ @mathuz Why did you change the question? The question, which you asked is proven. See my answer. If you want to ask another question, open a new topic. I made rollback. $\endgroup$ Commented Oct 2 at 15:02
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    $\begingroup$ @mathuz So open another topic with your new question. $\endgroup$ Commented Oct 2 at 15:21
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    $\begingroup$ No more rolling it back, okay? $\endgroup$ Commented Oct 2 at 15:48
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    $\begingroup$ @Shaun: I do not think that the question has been changed. The initial version already explained $\lesssim_a$ as “for each $a\in (0,1)$ we can find a uniform constant $C_a>0$ such that the above inequality holds for all $x>0$.” In revision #3 this was rewritten as $\cdots \leq C_a \cdots $ in an attempt to make the question clearer. Which means that the original question has not yet been answered. $\endgroup$ Commented Oct 2 at 17:48
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    $\begingroup$ @Shaun: I would suggest to rollback to revision #3: That is what OP meant from the beginning, without using the special notation $\lesssim_a$. But I do not want to start another rollback war. $\endgroup$ Commented Oct 2 at 18:42

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Let $f(x)=\frac{1}{(1+x)^{a}}-\frac{\ln(1+x)}{x}.$

We see that $$\lim_{x\rightarrow0^+}f(x)=0,$$ but $$\lim_{x\rightarrow0^+}f'(x)=\lim_{x\rightarrow0^+}\left(-\frac{a}{(1+x)^{1+a}}-\frac{1}{x(1+x)}+\frac{\ln(1+x)}{x^2}\right)=$$ $$=\lim_{x\rightarrow0^+}\frac{(1+x)\ln(1+x)-x}{x^2(1+x)}-a=\lim_{x\rightarrow0^+}\frac{\ln(1+x)}{2x+3x^2}-a=$$ $$=\lim_{x\rightarrow0^+}\frac{\frac{1}{1+x}}{2+6x}-a=\frac{1}{2}-a\geq0,$$ which says that for any $a>\frac{1}{2}$ the inequality is wrong.

Let $a=\frac{1}{2}$.

We need to prove that $g(x)\geq0,$ where $$g(x)=\frac{x}{\sqrt{x+1}}-\ln(x+1).$$

Indeed, $$g'(x)=\frac{1}{\sqrt{x+1}}-\frac{x}{2\sqrt{(x+1)^3}}-\frac{1}{x+1}=$$ $$=\frac{2(x+1)-x-2\sqrt{x+1}}{2\sqrt{(x+1)^3}}=\frac{\left(\sqrt{x+1}-1\right)^2}{2\sqrt{(x+1)^3}}\geq0,$$ which says $$g(x)\geq g(0)=0$$ and the inequality is proven for $a=\frac{1}{2}.$

Let $0\leq a<\frac{1}{2}.$

Thus, $$\frac{x}{(1+x)^a}-\ln(x+1)\geq\frac{x}{(1+x)^{\frac{1}{2}}}-\ln(x+1)\geq0,$$ which finishes the solution.

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  • $\begingroup$ Got it, thanks. I think it holds for all $a<1$ up to some constant $C_a$. We actually want to show that $\log (1+x) \le C_a \frac{x}{(1+x)^a}$ for some constant $C_a$. $\endgroup$ Commented Oct 2 at 13:22
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    $\begingroup$ @mathuz You are welcome! Your second question it's really another problem... $\endgroup$ Commented Oct 2 at 13:25

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