Let some (fixed) real $k >0$ and positive reals $a,b,c$. Consider the conjecture $$ \left(\frac{a}{b+c}\right)^k +\left(\frac{b}{a+c}\right)^k+\left(\frac{c}{a+b}\right)^k \geq \min \{\frac{3}{2^k} ; 2 \} $$ Some cases are known: $k = \frac12$ gives the bound 2, which is proved here.

$k = 1$ gives the bound 1.5, which is Nesbitt's inequality

I guess $k = 2$ should be known but I couldn't find it.

Does the conjecture hold?


1.$\frac{3}{2^k}\leq2$ or $k\geq\log_23-1=0.5849...$.

Now, let $a+b+c=3$.

Thus, we need to prove that $\sum\limits_{cyc}f(a)\geq\frac{3}{2^k}$, where $$f(x)=\left(\frac{x}{3-x}\right)^k.$$ Indeed, $$f''(x)=\frac{3kx^{k-2}(2x+3k-3)}{(3-x)^{k+2}}>0$$ for all $1\leq x<3,$ which says that by Vasc's RCF Theorem it's enough to check our inequality for an equality case of two variables, which is smooth enough.

About RCF Theorem see here:


  1. $0<k\leq\frac{1}{2}.$

We need to prove that $$\sum_{cyc}\left(\frac{a}{b+c}\right)^k\geq2.$$ Indeed, let $a^k=\sqrt{x},$ $b^k=\sqrt{y}$ and $c^k=\sqrt{z}.$

Thus, since $g(x)=x^{\frac{1}{2k}}$ is a convex function, by Karamata we obtain: $$(y+z)^{\frac{1}{2k}}+0\geq y^{\frac{1}{2k}}+z^{\frac{1}{2k}},$$ which gives $$\sum_{cyc}\left(\frac{a}{b+c}\right)^k=\sum_{cyc}\frac{\sqrt{x}}{\left(y^{\frac{1}{2k}}+z^{\frac{1}{2k}}\right)^k}\geq\sum_{cyc}\sqrt{\frac{x}{y+z}}\geq2.$$

  1. $\frac{1}{2}<k<\log_2\frac{3}{2}.$

In this case I have no a nice proof.

Since $f$ has on $(0,3)$ an unique inflection point for $x=x_0=\frac{3}{2}(1-k)<1,$

it's impossible that $\{a,b,c\}\subset\left(0,x_0\right).$

Now, let $a\leq b\leq c$.

We have three cases.

a) $a\leq b\leq x_0\leq c.$

In this case we can use Karamata and we'll get an inequality of one variable.

b) $a\leq x_0\leq b\leq c$.

In this case we can use Jensen and we'll get an inequality of one variable.

c) $x_0\leq a\leq b\leq c$.

In this case our inequality is obviously true by Jensen.

  • $\begingroup$ You consider the square root ($k = \frac12$)in your $f(x)$, How does this generalize to other $k$, as asked in the conjecture ? $\endgroup$ – Andreas Feb 1 '19 at 19:30
  • $\begingroup$ @Andreas Sorry. I fixed. $\endgroup$ – Michael Rozenberg Feb 1 '19 at 20:55

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