# Prove that $f(1)\leq f(x)<f(0)$ and another conjecture .

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

Let $$0 a real number and $$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}}

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))

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 .

## Update

Case $$n=1$$

One can prove that the function :

$$h(x)=(1+x)^{\frac{-1}{x}}$$

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 :

$$h^{-1}(x)=\frac{\operatorname{W}(x\log(x))-\log(x)}{\log(x)}$$

With the Lambert's function .

If you have a nice way to solve it .

Thanks you very much .

• I think that in the conjeture is possible used Jensen's inequality... Aug 2, 2020 at 16:46
• @AsdrubalBeltran I'm aware of that but it's simpler using first derivative. Aug 3, 2020 at 8:11
• 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 $f(x)$. Aug 9, 2020 at 11:47
• @Digitallis it's more a limit case it's symbolic . Aug 9, 2020 at 12:03

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.

• 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). Aug 16, 2020 at 8:30