# Generalization of Nesbitts's inequality

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?

• – Dr. Sonnhard Graubner Feb 1 '19 at 18:19
• There is a very nice proof of your inequality for $k=\frac{27}{46}.$ – Michael Rozenberg Feb 1 '19 at 18:31
• @MichaelRozenberg "There is ... " ? – Andreas Feb 1 '19 at 19:31
• Yes, I found this proof. – Michael Rozenberg Feb 1 '19 at 20:44

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.

https://pdfs.semanticscholar.org/1419/e0baa6b073c5903e430ddb0c9a154c61d208.pdf

1. $$0

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}

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.

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