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following this kind of inequality One of my old inequality (very sharp) I propose this because I don't see it on the forum :

Let $a,b,c>0$ and $a+b+c=1$ with $r\in(\frac{1}{2},1)$ and $a\geq b \geq c$ then we have : $$\frac{a}{a^r+b^r}+\frac{b}{b^r+c^r}+\frac{c}{c^r+a^r}\geq \frac{a}{a^r+c^r}+\frac{c}{c^r+b^r}+\frac{b}{b^r+a^r}$$

First of all it's a conjecture where I don't find counter-examples . Secondly when $r\in(0,\frac{1}{2})$ the inequality is reversed .I use Pari-gp for that .Furthermore (if it's true) I think it's really not new so I add the tag reference request.We have an equality case when $r=0.5$ whenever $a,b,c>0$.

So if you have idea to prove it or disprove it...

Thanks a lot .

Michael Rozenberg
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Miss and Mister cassoulet char
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1 Answers1

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If $\prod\limits_{cyc}(a-b)=0$, so it's obvious.

Let $a>b>c.$

Thus, we need to prove that: $$\sum_{cyc}\left(\frac{a}{a^r+b^r}-\frac{a}{a^r+c^r}\right)\geq0$$ or $$\sum_{cyc}\frac{a(c^r-b^r)}{(a^r+b^r)(a^r+c^r)}\geq0$$ or $$\sum_{cyc}a(c^r-b^r)(b^r+c^r)\geq0$$ or $$\sum_{cyc}a(c^{2r}-b^{2r})\geq0$$ or $$a^{2r}(b-c)+c^{2r}(a-b)-b^{2r}(a-b+b-c)\geq0$$ or $$\frac{a^{2r}-b^{2r}}{a-b}\geq\frac{b^{2r}-c^{2r}}{b-c}.$$ Now, use the Lagrange's mean value theorem for $f(x)=x^{2r}$ and that $f'$ increases.

Michael Rozenberg
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