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Conjecture, Prove that :

$$\sum_{cyc}\frac{a}{a^n+1}\leq \sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$$ Under the assumptions $a\geq b\geq 1\geq c>0$ such that $abc=1$ and $\frac{c}{c^n+1}\geq \frac{b}{b^n+1}\geq \frac{a}{a^n+1}$ and $\frac{a}{a+b}\geq \frac{b}{b+c}\geq \frac{c}{c+a}$ and finally $n\geq 10$ a natural number.

My work

We start with the following expression :

$$\Big(\sum_{cyc}\frac{a}{a^n+1}\Big)\Big(\sum_{cyc}\frac{a}{a+b}\Big)$$

Using Tchebytchev's inequality and the order we have :

$$\Big(\sum_{cyc}\frac{a}{a^n+1}\Big)\Big(\sum_{cyc}\frac{a}{a+b}\Big)\leq 3\Big(\frac{a}{a^n+1}\frac{c}{c+a}+\frac{b}{b^n+1}\frac{b}{b+c}+\frac{a}{a+b}\frac{c}{c^n+1}\Big)$$

Now we study the following expression :

$$3\Big(\frac{ac+a^n}{(a^n+1)(c+a)}+\frac{b^2+b^{n-1}c}{(b^n+1)(b+c)}+\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}\Big)$$

Take one element like :

$$\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}$$

We prove that :

$$\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}\leq \frac{c}{c^2+1}\leq \frac{1}{2}$$

Or :

$$\frac{a+c^{n-2}b}{(a+b)(c^n+1)}\leq \frac{1}{c^2+1}$$

Or :

$$\frac{(a+c^{n-2}b)(c^2+1)}{(a+b)(c^n+1)}\leq 1$$

Or :

$$(a+c^{n-2}b)(c^2+1)\leq (a+b)(c^n+1)$$

Or :

$$ac^2+a+c^nb+c^{n-2}b\leq ac^n+a+bc^n+b$$

Or : $$ac^2+c^{n-2}b\leq ac^n+b$$

Using the fact that $abc=1$ we have :

$$ac^2+\frac{c^{n-2}}{ac}\leq ac^n+\frac{1}{ac}$$

Or :

$$a^2c^3+c^{n-2}\leq a^2c^{n+1}+1$$

Wich have the form :

$$x+y\leq xy+1$$

Wich can be solved with tangent hyperbolic .

Simpler if we have $a+c\geq 2$ and $ac\leq 1$ we deduce that :

$$\frac{ac+a^n}{(a^n+1)(c+a)}\leq \frac{1}{2}$$

Similar method conducts to :

$$\frac{b^2+b^{n-1}c}{(b^n+1)(b+c)}\leq \frac{1}{2}$$

Summing each elements we have :

$$3\Big(\frac{ac+a^n}{(a^n+1)(c+a)}+\frac{b^2+b^{n-1}c}{(b^n+1)(b+c)}+\frac{ac+c^{n-1}b}{(a+b)(c^n+1)}\Big)\leq \frac{9}{4}$$

But :

$$\frac{a^n}{(a^n+1)(c+a)}=\frac{\frac{1}{a}\frac{1}{c}}{(\frac{1}{a^n}+1)(\frac{1}{a}+\frac{1}{c})}$$

The same trick apply to the other elements conducts to :

$$3\Big(\frac{a}{a^n+1}\frac{c}{c+a}+\frac{b}{b^n+1}\frac{b}{b+c}+\frac{a}{a+b}\frac{c}{c^n+1}\Big)\leq \frac{9}{4}$$

And we deduce that :

$$\Big(\sum_{cyc}\frac{a}{a^n+1}\Big)\Big(\sum_{cyc}\frac{a}{a+b}\Big)\leq \frac{9}{4}$$

But with the assumptions we have : $$\sum_{cyc}\frac{a}{a+b}\geq \frac{3}{2}$$

And so :

$$\sum_{cyc}\frac{a}{a^n+1}\leq \frac{3}{2}$$

My questions

Can someone correct it if it's wrong ?

With all this elements of proof can someone achieve or complete my proof ?

Have you other method to learn ?

Any helps is greatly appreciated .

Thanks in advance !

Ps : if you have a counter-example say where I'm wrong in my proof please .

PPs:I add the tag "contest-maths" to see if there is an elegant hand proof other than mine .

Update :

My trick is false because it works only for $a=b=c=1$ so the idea is to prove

$$3\Big(\frac{ac-a^n}{(a^n+1)(c+a)}+\frac{b^2-b^{n-1}c}{(b^n+1)(b+c)}+\frac{ac-c^{n-1}b}{(a+b)(c^n+1)}\Big)\leq 0$$

Under some assumptions .

The idea is to use the method A special case of Karamata's inequality to solve one or more Olympiad inequality? . I will develop it later but with this method it's easy the only problem is the order .

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I will prove the second inequality $\sum_{cyc}\frac{a}{a^2+1}\leq \frac{3}{2}$ for you.
Note that for $x\gt0,$ $$\dfrac{x}{x^2+1}=\left(x+\dfrac1x\right)^{-1}$$ and by AM-GM inequality $$x+\dfrac1x\ge2$$ with equality occurs at $x=1.$ Hence the result follows.

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  • $\begingroup$ You can find two different ways to prove the same result from here. $\endgroup$ – Bumblebee Jul 11 at 16:31

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