It's a problem found with the help of WA .

Let $0<x$ a real number and $n\geq 1$ a natural number then we have : $$ f(1)\leq f(x)=(1+x)^{\frac{-1}{(nx)}}+\Big(1+\frac{1}{x}\Big)^{-\frac{x}{n}}<f(0)$$

I have also conjectured that :

Let $1\leq x$ a real number and $n\geq 1$ a natural number then we have : $$f\Big(\frac{x+\frac{1}{x}}{2}\Big)\leq f(x)$$

This conjecture I think is useful because of this fact :

Let $g(x)=\frac{x+\frac{1}{x}}{2}$ and $x\geq 1$ a real number then we have :

$$f(1) \leq f(g_n(x))\leq f(g_{n-1}(x))\leq \cdots \leq f(g(x))\leq f(x)\leq f(g^{-1}(x))\leq \cdots\leq f(g^{-1}_n(x))<f(0)$$

Where we speak about the inverse of the function $g(x)$ and the iteration ($n$ to $n$ times) of the function $g(x)$ with itself .

So the idea is to prove more generaly that $x=1$ is a minimum and $x=0$ is an infimum .

Well I have tried the same method as here by user Robin Aldabanx (first answer with bounty) for the first case or $n=1$ .

I have tried power series as well without success this time .

I have been also inspired by the Polya's proof of Am-Gm but no good issues .


Case $n=1$

One can prove that the function :


is concave on $(0,\infty)$ . So we can apply Karamata's inequality and a majorization to get something of the kind :

$$h(x)+h\Big(\frac{1}{x}\Big)\geq h(y)+h\Big(\frac{1}{y}\Big)\quad (1)$$

The inequality $(1)$ gives information on $f(x)$ in the case where $n=1$.Via the majorization we can say as it's increasing or decreasing .

Moreover I think we can apply this method to the general case $n\geq 1$.

I don't know if it's really relevant but the inverse function of $h(x)$ is :


With the Lambert's function .

If you have a nice way to solve it .

Thanks you very much .

  • $\begingroup$ I think that in the conjeture is possible used Jensen's inequality... $\endgroup$ Aug 2, 2020 at 16:46
  • $\begingroup$ @AsdrubalBeltran I'm aware of that but it's simpler using first derivative. $\endgroup$ Aug 3, 2020 at 8:11
  • $\begingroup$ What do you mean by $f(0)$ here ? You've only defined $f(x)$ for $x>0$ and there is a possible division by $0$ in the exponent of the first term in $f(x)$. $\endgroup$
    – Digitallis
    Aug 9, 2020 at 11:47
  • $\begingroup$ @Digitallis it's more a limit case it's symbolic . $\endgroup$ Aug 9, 2020 at 12:03

1 Answer 1


Half of half an answer:

By the convexity of $x ↦ x^{-1/n}$, we have

$$f(x) = (1 + x)^{-1/nx} + \left(1 + \frac1x\right)^{-nx} \geq 2\left(\frac{(1 + x)^{1/x} + \left(1 + \frac1x\right)^{x}}2\right)^{-1/n}$$

So, if you can show that the maximum of the bracket is what you get when you input $x=1$ (as a plot confirms):

$$ \max_{x>0} \left((1 + x)^{1/x} + \left(1 + \frac1x\right)^{x}\right) = 4,$$

then you get the required lower bound for each $n$:

$$f(x) \geq 2 (2)^{-1/n} = f(1),$$

That is, the lower bound reduces to something like the $n=1$ case, which you seem happy with.

  • $\begingroup$ First thanks it seems we can also apply a Karamata-like inequality with the function of the bracket .If we take $x^{\frac{1}{n}}$ we have the RHS but I don't check if it's good .As it helps me (+1). $\endgroup$ Aug 16, 2020 at 8:30

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