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Here is a known double inequality for positive numbers: $$ 1\le\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\le 2 $$ Source: https://www.cut-the-knot.org/m/Algebra/Crux4196.shtml

I'm curious if at least one of them can be generalised in the following way: $$ C_1(a^{r-1}+b^{r-1}+c^{r-1})\le\frac{a^r}{a+b}+\frac{b^r}{b+c}+\frac{c^r}{c+a}\le C_2(a^{r-1}+b^{r-1}+c^{r-1}) $$ The inequality (or at least one of them, if not possible for both) should be true for any reals $a,b,c>0$, $r\ge 1$; $C_1,C_2$ are positive numbers not depending on $a,b,c,r$.

I searched a lot, on this site and others like sets of cyclic inequalities e.g. https://www.cut-the-knot.org/m/Algebra/CyclicInequality1.shtml or papers e.g. http://www.m-hikari.com/imf/imf-2018/1-4-2018/p/laiIMF1-4-2018.pdf, but didn't found such result. The only related thing I found is $ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \geq 1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$, a stronger version of given inequality, found here: Cyclic Inequality in n (at least 4) variables

Michael Rozenberg
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user807920
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  • If you do not set fixed values (analogous to $1$ and $2$ in the original double bounding) but put them as functions $C_1$ and $C_2$ to be determined, I think it's almost natural that in fact these functions do exist for each value of the exponent $r$.Another problem is the determination of these functions. – Ataulfo Jul 11 '20 at 21:32

1 Answers1

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The left inequality.

By Chebyshov we obtain: $$\sum_{cyc}\frac{a^r}{a+b}>\sum_{cyc}\frac{a^r}{a+b+c}\geq\frac{a^{r-1}+b^{r-1}+c^{r-1}}{3}.$$ Id est, we can say that $C_1=\frac{1}{3}.$

Also, for $r=\frac{7}{6}$ by Buffalo Way we can show that the following inequality is true. $$\sum_{cyc}\frac{a^r}{a+b}\geq\frac{a^{r-1}+b^{r-1}+c^{r-1}}{2}.$$ For $r=\frac{8}{7}$ the last inequality is wrong already.

The right inequality. $$\sum_{cyc}\frac{a^r}{a+b}<\sum_{cyc}\frac{a^r}{a}=a^{r-1}+b^{r-1}+c^{r-1}$$ and we can say that $C_2=1$.

Michael Rozenberg
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