Find $k$ such that $\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}$ for $xyz=1$.

Question

Find all $k\in\mathbb{R}^+$ such that for all $xyz=1$, $x,y,z\in\mathbb{R}^+$, we have $$\frac{1}{(1+x)^k}+\frac{1}{(1+y)^k}+\frac{1}{(1+z)^k}\geq\frac{3}{2^k}.$$

If we set $$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a},$$ where $a,b,c\in\mathbb{R}^+$, then the inequality is equivalent to $$\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\geq\frac{3}{2^k}.$$ For $k=1$ this is obviously false, as we take $b\to0$ and $c\to\infty$. In an olympiad test I saw that the case $k=\sqrt3$ is true (though I do not know how to prove that). I then thought that maybe this is true for all $k>1$. Am I correct and how do I prove it?

Answer 1

I think the answer is $k_{min}=\log_23$.

Let $x=e^a$, $y=e^b$, $z=e^c$ and $f(x)=\frac{1}{(1+e^x)^k}$.

Hence, $a+b+c=0$ and we need to prove that

$$\sum_{cyc}f(a)\geq3f\left(\frac{a+b+c}{3}\right)$$ But $f''(x)=\frac{2ke^x(ke^x-1)}{(1+e^x)^{k+2}}\geq0$ for all $k\geq1$ and $x\geq0$.

Thus, by Vasc's RCF Theorem for all $k\geq1$ it's enough to prove our inequality

for $y=x$ and $z=\frac{1}{x^2}$, which gives $\frac{2}{(1+x)^k}+\frac{1}{\left(1+\frac{1}{x^2}\right)^k}\geq\frac{3}{2^k}$ and for $x\rightarrow+\infty$ we get $k\geq\log_23$.

Answer 2

As shown below, an easy partial result is that $k_{\text{min}} \ge 1$.

Let $k$ be fixed with $0 < k < 1$.

WIth the substitutions

$$x=\frac{a}{b},\,y=\frac{b}{c},\,z=\frac{c}{a}$$

let $a = b = 1$, and let $c \to 0^{+}$. Then

\begin{align*} \;&\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\\[6pt] &\to \;\frac{1}{2^k} + 1 + 0\\[6pt] &=\;\frac{1 + 2^k}{2^k}\\[6pt] &<\;\frac{1 + 2}{2^k}\\[6pt] &=\;\frac{3}{2^k}\\[6pt] \end{align*}

It follows that $k_{\text{min}} \ge 1$, as claimed.

However, as shown below, the reasoning by which the OP (Yuxiao Xie) excluded the value $k=1$ is not correct.

Suppose $k = 1$. Then as$\;\,b \to 0^{+}$ and $\,c \to \infty$,

\begin{align*} \;&\frac{a^k}{(a+b)^k}+\frac{b^k}{(b+c)^k}+\frac{c^k}{(c+a)^k}\\[6pt] &\to\;1+ 0 + 1\\[6pt] &=\;2\\[6pt] &>\;\frac{3}{2}\\[6pt] \end{align*}

which doesn't break the required inequality, hence the value $k=1$ is not automatically excluded.

In fact, I suspect that $k_{\text{min}} = 1$.

Update: It's not true, as I had previously thought, that $k_{\text{min}} = 1$.

As a simple counterexample, let $(x,y,z) = \left(\frac{1}{8},2,4\right)$. Then

$$\frac{1}{(1+x)}+\frac{1}{(1+y)}+\frac{1}{(1+z)} = \frac{64}{45} < \frac{3}{2}$$