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Let $f(x)=(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x, 0<x\leq 1.$ Prove that $f$ is strictly increasing and $e<f(x)\leq 4.$

In order to study the Monotonicity of $f$, let $$g(x)=\log f(x)=\frac{1}{x}\log (1+x)+x\log \left(1+\frac{1}{x}\right).$$ And $f$ and $g$ has the same Monotonicity. By computation, $$g'(x)=\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}.$$ As we know $\frac{x}{1+x}-\log (1+x)\leq 0$ and $\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq 0$. So it does not determine the sign of $g'(x)$. If we compute the second derivative $g''(x)$, you will find it is also difficult to determine the sign of $g''(x)$. Our main goal is to prove $$\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}>0.$$

Is there some tricks to prove this result. Any help and hint will welcome.

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    $\begingroup$ I believe that the solution to this problem is to use the symmetry of function $f(x)$. The symmetry is $f(x)=f(1/x)$. $\endgroup$ – MathOverview Sep 7 '18 at 12:11
  • $\begingroup$ Just an observation: If you manage to at least show that $f'(x) \not= 0$ for all $x \in (0,1]$, then you are done because $\lim_{x \to 0^+} f(x) = e < 4 = f(1),$ so there is no other option for $f$ other than to be strictly increasing on $(0, 1]$. $\endgroup$ – MisterRiemann Sep 7 '18 at 12:15
  • $\begingroup$ Is this related to math.stackexchange.com/q/2899031/42969, or is the resemblance coincidental? $\endgroup$ – Martin R Sep 7 '18 at 12:22
  • $\begingroup$ @ Martin R: thank you for your hint of math.stackexchange.com/q/2899031/42969 Actually, the question I asked is confused me for several weeks, and I have seen math.stackexchange.com/q/2899031/42969 before, and then there is only one answers. And now I see there are several answers. And one answer use the result I have posted. It is resemblance coincidental. $\endgroup$ – Riemann Sep 7 '18 at 12:26
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The right inequality.

We can use the TL method here.

We need to prove that $$(1+a)^b(1+b)^a\leq4$$ for $a>0$, $b>0$ such that $ab=1$, which is $$\frac{\ln(1+a)}{a}+\frac{\ln(1+b)}{b}\leq2\ln2.$$ But $$\sum_{cyc}\left(\ln2-\frac{\ln(1+a)}{a}\right)=\sum_{cyc}f(a),$$ where $f(a)=\ln2-\frac{\ln(1+a)}{a}-(\ln2-0.5)\ln a$.

Easy to show that $f(a)\geq0$ for all $0<a\leq11$.

But for $a>11$ we obtain: $$(1+a)^b(1+b)^a=(1+a)^{\frac{1}{a}}\left(1+\frac{1}{a}\right)^a<(1+11)^{\frac{1}{11}}e<4.$$

The left inequality.

We need to prove that: $$\frac{1}{x}\ln(1+x)+x\ln\left(1+\frac{1}{x}\right)>1$$ or $g(x)>0$, where $$g(x)=(1+x^2)\ln(1+x)-x^2\ln{x}-x.$$ Indeed, $$g'(x)=2x\ln(1+x)+\frac{1+x^2}{1+x}-2x\ln{x}-x-1=2x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)>0$$ because $$\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)'=-\frac{1}{x(1+x)^2}<0$$ and $$\ln\left(1+\frac{1}{1}\right)-\frac{1}{1+1}>0.$$ Id est, $$g(x)>\lim_{x\rightarrow0^+}g(x)=0$$ and we proved that $e<f(x)\leq4.$

Now, we'll prove that $f$ increases on $(0,1].$

By your work we need to prove that $$\frac{1}{x^2}\left(\frac{x}{1+x}-\ln(1+x)\right)+\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq0$$ for all $0<x\leq1$ or $$(x^2-1)\ln(1+x)-x^2\ln{x}+\frac{(1-x)x}{1+x}\geq0$$ and since $$\ln{x}\leq\frac{2(x-1)}{1+x},$$ it's enough to prove that $$-(1-x^2)\ln(1+x)+\frac{2x^2(1-x)}{1+x}+\frac{(1-x)x}{1+x}\geq0$$ or $$\ln(1+x)\leq\frac{x(2x+1)}{(1+x)^2},$$ which is smooth.

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Using $$\dfrac{x}{x+1}<x\ln\left(1+\dfrac1x\right)<1$$ and for $x\to\dfrac1x$ $$\dfrac{1}{x+1}<\dfrac1x\ln\left(1+x\right)<1$$ then $$\dfrac1x\ln\left(1+x\right)+x\ln\left(1+\dfrac1x\right)>1$$

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  • $\begingroup$ You trick is very nice.Can you give some help for the fonotonicity of the function $f(x)$.Thank you ! $\endgroup$ – Riemann Sep 7 '18 at 14:37
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Michael Rozenberg's answer has it all.

Here are two remaining proofs in Michael Rozenberg's answer:

  1. Show: $\ln{x}\leq\frac{2(x-1)}{1+x}$

We have $\ln(1+y)\leq\frac{2y}{{2+y}}$ for $-1<y<0$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), i.e. $\ln(x)\leq\frac{2(x-1)}{{1+x}}$ for $0<x<1$. This is exactly what needs to be shown.

  1. Show: $\ln(1+x)\leq\frac{x(2x+1)}{(1+x)^2}$

We have $\ln(1+x)\leq\frac{x}{\sqrt{1+x}}$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), so we may prove

$\frac{1}{\sqrt{1+x}}\leq\frac{2x+1}{(1+x)^2}$ or $(1+x)^3-{(1+2x)^2}\leq 0$ or $x^2 - x - 1<0$ or $x( 1-x) + 1>0$ which is true since $1-x \ge 0$.

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  • $\begingroup$ Sorry @Andreas , I have made a mistake. $g'(x)$ should be $g'(x)=\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}.$ $\endgroup$ – Riemann Sep 7 '18 at 13:11

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