# Pretty conjecture $x^{\left(\frac{y}{x}\right)^n}+y^{\left(\frac{x}{y}\right)^n}\leq 1$

inspired (again) by an inequality of Vasile Cirtoaje I propose my own conjecture :

Let $$x,y>0$$ such that $$x+y=1$$ and $$n\geq 1$$ a natural number then we have : $$x^{\left(\frac{y}{x}\right)^n}+y^{\left(\frac{x}{y}\right)^n}\leq 1$$

First I find it very nice because all the coefficient are $$1$$ .

I have tested with Geogebra until $$n=50$$ without any counter-examples.

Furthermore we have an equality case as $$x=y=0.5$$ or $$x=1$$ and $$y=0$$ and vice versa .

To solve it I have tried all the ideas here

My main idea was to make a link with this inequality (my inspiration) see here

So if you can help me to solve it or give me an approach...

...Thanks for all your contributions !

## Little update

I think there is also an invariance as in question here Conjecture $a^{(\frac{a}{b})^p}+b^{(\frac{b}{a})^p}+c\geq 1$

## Theoretical method

Well,Well this method is very simple but the result is a little bit crazy (for me (and you ?))

Well ,I know that if we put $$n=2$$ we can find (using parabola) an upper bound like

$$x^{\left(\frac{1-x}{x}\right)^2}\leq ax^2+bx+c=p(x)$$ And $$(1-x)^{\left(\frac{x}{1-x}\right)^2}\leq ux^2+vx+w=q(x)$$

on $$[\alpha,\frac{1}{2}]$$ with $$\alpha>0$$ and such that $$p(x)+q(x)<1$$

In the neightborhood of $$0$$ we can use a cubic .

Well,now we have (summing) :

$$x^{\left(\frac{1-x}{x}\right)^2}+(1-x)^{\left(\frac{x}{1-x}\right)^2}\leq p(x)+q(x)$$

We add a variable $$\varepsilon$$ such that $$(p(x)+\varepsilon)+q(x)=1$$

Now we want an inequality of the kind ($$k\geq 2$$):

$$x^{\left(\frac{1-x}{x}\right)^{2k}}+(1-x)^{\left(\frac{x}{1-x}\right)^{2k}}\leq (p(x)+\varepsilon)^{\left(\frac{1-x}{x}\right)^{2k-2}}+q(x)^{\left(\frac{x}{1-x}\right)^{2k-2}}$$

Now and it's a crucial idea we want something like :

$$\left(\frac{x}{1-x}\right)^{2k-2}\geq \left(\frac{1-(p(x)+\varepsilon)}{q(x)}\right)^y$$

AND :

$$\left(\frac{1-x}{x}\right)^{2k-2}\geq \left(\frac{1-q(x)}{p(x)+\varepsilon}\right)^y$$

Now it's not hard to find a such $$y$$ using logarithm .

We get someting like :

$$x^{\left(\frac{1-x}{x}\right)^{2k}}+(1-x)^{\left(\frac{x}{1-x}\right)^{2k}}\leq q(x)^{\left(\frac{1-q(x)}{q(x)}\right)^{y}}+(1-q(x))^{\left(\frac{q(x)}{1-q(x)}\right)^{y}}$$

Furthermore the successive iterations of this method conducts to $$1$$ because the values of the differents polynomials (wich are an approximation of the initial curve) tend to zero or one (as abscissa).

The extra-thing (and a little bit crazy) we can make an order on all the values.

## My second question

Is it unusable as theoretical\practical method ?

• Have you already tried the limit cases around 1 and 0.5? Jul 15, 2020 at 17:15
• @Biggus Dickus Python, Does this help you to prove your conjecture ? Jul 15, 2020 at 20:57
• @zeraoulia rafik interesting but it seems to be not sufficient to prove the entire inequality.Anyway with this I have a special case .(+1) for your question. Jul 17, 2020 at 13:37
• Oups this not real number but natural number ah ! Anyway have a good day. :-) Jul 17, 2020 at 13:43
• @ErikSatie: I would replace $n$ with a positive variable and try Method of Lagrange Multipliers to see if we can extract some thing. Dec 6, 2020 at 3:00

I am sorry for not proving your conjecture, but I thought of writing up my thoughts as it might help you.

Without loss of generality, we can say that $$y \geq 0.5$$. Let $$q=y/x$$ and because $$y \geq 0.5$$ we have $$q\geq 1$$ (you can do everything in reverse with $$x \geq 0.5$$ and $$q\leq 1$$). Let $$a_0=x$$ and $$b_0=y$$. Then we can write: $$x^{\left( \frac{y}{x} \right)^n} + y^{\left(\frac{x}{y}\right)^n}=x^{q^n}+y^{\left(\frac{1}{q}\right)^n}=a_n+b_n,$$ with \begin{align} a_n &= a_{n-1}^q,\\ b_n &= b_{n-1}^{1/q}. \end{align}

It is easy to verify that with $$b_n\geq0.5$$, also $$b_{n+1}\geq 0.5$$. Now, let us assume that $$a_{n-1} \leq 1-b_{n-1}$$. For $$n=1$$, we have $$a_0 \leq 1-b_0$$ (more specifically, $$a_0=1-b_0$$). Because of that, we can write (remember that $$q \geq 1$$): \begin{align} a_{n+1} + b_{n+1} &= a_n^q+b_n^{1/q} \\ &\leq (1-b_n)^q + b_n^{1/q}. \end{align}

What is left, is to prove that $$(1-b_n)^q+b_n^{1/q} \leq 1, \quad \forall \,\,\, b_n \in [0.5, 1], q\geq1$$

This is where I got stuck, but perhaps you know how to continue from here. In the figure below, I plotted $$(1-b)^q+b^{1/q}$$ with $$b$$ on the interval $$[0.5,1]$$ for various values of $$q$$ and it seems that the result is always equal or below $$1$$.

Let us prove the case when $$0 < x \le \frac{14}{33}$$.
By Bernoulli inequality, we have $$y^{(x/y)^n} = (1 - x)^{(x/y)^n} \le 1 - x(x/y)^n.$$ It suffices to prove that $$x^{(y/x)^n} + 1 - x(x/y)^n \le 1$$ or $$\ln x \le \frac{- \ln[(y/x)^n]}{(y/x)^n - 1}.$$ Since $$u \mapsto \frac{-\ln u}{u - 1}$$ is strictly increasing on $$u > 1$$, it suffices to prove that $$\ln x \le \frac{- \ln[(y/x)]}{(y/x) - 1}.$$ With the substitution $$x = \frac{1}{z}$$, it suffices to prove that, for all $$z \ge \frac{33}{14}$$, $$z\ln(z) - \ln(z - 1) - 2\ln z \ge 0$$ which is true (easy).