# Show that $\left(\frac{x_1^{x_2}}{x_2}\right)^p+\left(\frac{x_2^{x_3}}{x_3}\right)^p+\cdots+\left(\frac{x_n^{x_1}}{x_1}\right)^p\ge n$ for any $p\ge1$

The inequality $$\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$$ for all $$a,b>0$$ was shown here using first-order Padé approximants on each exponent, where the minimum is attained at $$a=b=1$$.

By empirical evidence, it appears that inequalities of this type hold for an arbitrary number of variables. We can phrase the generalised problem as follows.

Let $$(x_i)_{1\le i\le n}$$ be a sequence of positive real numbers. Define $$\boldsymbol a=\begin{pmatrix}a_1&\cdots&a_n\end{pmatrix}$$ such that $$a_k=x_k^{x_{k+1}}/x_{k+1}$$ for each $$1\le k and $$a_n=x_n^{x_1}/x_1$$. How do we show that $$\|\boldsymbol a\|_p^p\ge n$$ for any $$p\ge1$$?

As before, AM-GM is far too weak since the inequality $$\displaystyle\|\boldsymbol a\|_p^p\ge 2\left(\prod_{\text{cyc}}\frac{x_1^{x_2}}{x_2}\right)^{1/{2p}}$$ does not guarantee the result when at least one $$x_i$$ is smaller than $$1$$. We can eliminate the exponent on the denominator by taking $$x_i=X_i^{1/p}$$ so that $$\displaystyle\|\boldsymbol a\|_p^p=\sum_{\text{cyc}}\frac{X_1^{X_2^{1/p}}}{X_2}$$ but the approximant approach no longer becomes feasible; even in the case where $$p$$ is an integer the problem reduces to a posynomial inequality of rational degrees. Perhaps there are some obscure $$L^p$$-norm/Hölder-type identities of use but I'm at a loss in terms of finding references.

Empirical results: In the interval $$p\in[1,\infty)$$, Wolfram suggests that the minimum is $$n$$ (Notebook result) which is obtained when $$\boldsymbol a$$ is the vector of ones. However, we note that in the interval $$p\in(0,1)$$, the empirical minimum no longer displays this consistent behaviour as can be seen in this Notebook result. The sequence $$\approx(1.00,2.00,2.01,3.36,3.00,4.00)$$ appears to increase almost linearly every two values, but I cannot verify it for a larger number of variables due to instability in the working precision.

• Interesting, I actually was thinking of a different kind of generalization, but I like this one as well. Of course, the usual stuff like when does one term dominate $n$ etc. will give you some easy bounds where the inequality is obvious. Oct 16, 2021 at 10:38
• @TheSimpliFire It looks good. Why someone downvoted it? Oct 16, 2021 at 14:33
• A counterexample is $(x_1, x_2) = (1/2, 2)$ and $p=1/4$. Unless I made an error, the expression becomes $\approx 1.89 < 2$. Perhaps you need to add the restriction $p \ge 1$. Oct 18, 2021 at 11:56
• @MartinR What if $n\ge 3$? Oct 18, 2021 at 12:06
• @RiverLi: I only played a bit with the case $n=2$ so far. It could be that $p \ge 0.5$ is sufficient. Oct 18, 2021 at 12:08

## 4 Answers

### Proof for $$p ≥ 1$$

Since $$u^p - 1 ≥ p(u - 1)$$ for all $$u ≥ 0$$, it suffices to prove the result for $$p = 1$$. That follows from

$$\frac{x^y}{y} - 1 ≥ \frac{1 + y \ln x}{y} - 1 = \ln x + \frac1y - 1 ≥ \ln x + \ln \frac1y = \ln x - \ln y$$

by cyclic summation over $$(x, y) = (x_i, x_{i + 1})$$.

### Conjectured proof for $$p ≥ \frac12$$

Since $$u^p - 1 ≥ 2p(u^{\frac12} - 1)$$ for all $$u ≥ 0$$, it suffices to prove the result for $$p = \frac12$$. Numerical evidence suggests that

$$\left(\frac{x^y}{y}\right)^{\frac12} - 1 ≥ \frac{\ln x}{2\sqrt[4]{1 + \frac13 \ln^2 x}} - \frac{\ln y}{2\sqrt[4]{1 + \frac13 \ln^2 y}}$$

for all $$x, y > 0$$. If this is true, cyclic summation yields the desired result.

### Counterexample for $$0 < p < \frac12$$

Let $$g(x) = \left(\frac{x^{1/x}}{1/x}\right)^p + \left(\frac{(1/x)^x}{x}\right)^p$$. Then $$g(1) = 2$$, $$g'(1) = 0$$, and $$g''(1) = 4p(2p - 1) < 0$$, so we have $$g(x) < 2$$ for $$x$$ in some neighborhood of $$1$$. This yields counterexamples for all even $$n$$:

$$\left(x, \frac1x, x, \frac1x, \dotsc, x, \frac1x\right), \quad x ≈ 1, \quad 0 < p < \frac12.$$

For $$n = 3$$, the best counterexample seems to be

$$(0.41398215, 0.73186577, 4.77292996), \quad 0 < p < 0.39158477.$$

• Let $x=e^X,y=e^Y$ so$$e^{(Xe^Y-Y)/2}\ge1+\frac X{2(1+X^2/3)^{1/4}}-\frac Y{2(1+Y^2/3)^{1/4}}$$For $X\ge0$ it suffices that$$e^{(Xe^Y-Y)/2}\ge1+\frac X2-\frac Y{2(1+Y^2/3)^{1/4}}$$As $\arg\min_Xe^{(Xe^Y-Y)/2}-X/2=-Ye^{-Y}$ it suffices that$$(Y+2)e^{-Y}+\frac Y{(1+Y^2/3)^{1/4}}\ge2$$Equality occurs at $Y=0$ so it suffices to show LHS has derivative $>0$, or$$Y-\log(1+Y)-\frac54\log(3+Y^2)+\log(6+Y^2)>\log\frac2{3^{1/4}}$$Equality occurs at $Y=0$ so it suffices to show LHS has derivative $>0$, or$$1-\frac1{1+Y}-\frac{5Y}{2(3+Y^2)}+\frac{2Y}{6+Y^2}>0$$or $2Y^4-Y^3+17Y^2-18Y+18>0$ which is true. Nov 2, 2021 at 11:17
• @Hans I am using $u^p - 1 ≥ p(u - 1)$ to observe that the result for $p = 1$ implies the result for $p ≥ 1$. Nov 3, 2021 at 0:32
• I don't know why 100 reputation is given to my answer. It should be given to Anders Kaseorg's excellent answer. Nov 3, 2021 at 16:54
• @RiverLi I agree; I suspect ErikSatie wanted to balance out the reputation. But thanks for starting the bounty, this answer totally deserves it. Nov 3, 2021 at 17:21
• I am very much intrigued by what motivated you to find the intermediate bounds $\frac{x^y}{y} - 1 ≥ \ln x - \ln y$ and especially $\left(\frac{x^y}{y}\right)^{\frac12} - 1 ≥ \frac{\ln x}{2\sqrt[4]{1 + \frac13 \ln^2 x}} - \frac{\ln y}{2\sqrt[4]{1 + \frac13 \ln^2 y}}$. Could you please shed some light on that?
– Hans
Nov 4, 2021 at 0:01

We prove the result by first deriving a useful inequality for the function $$(x,y)\mapsto (x^y/y)^p$$.

It is clear that $$(x^y/y)^p \ge \left(\min_{\alpha \in \mathbb{R}_+} (\alpha x)^{\alpha y}/(\alpha y)\right)^p.\tag{1}$$ For fixed $$x$$ and $$y$$, we can find the minimum of the function $$\alpha\mapsto (\alpha x)^{\alpha y}/(\alpha y)$$ by taking derivatives and setting to $$0$$. Doing this (e.g., in Mathematica) shows that the minimum is achieved by $$\alpha^\star = 1/({yW(ex/y)})$$, where $$W$$ is the Lambert W function. Plugged into $$(1)$$, this gives $$(x^y/y)^p \ge \left[\left(\frac{x/y}{W(ex/y)}\right)^{1/W(e x/y)}W(ex/y)\right]^p$$ Note that the right hand side is a function of the ratio $$x/y$$ only. It will be convenient to write it as a function of the logarithm of $$x/y$$, $$(x^y/y)^p\ge f(\ln x/y)$$ where $$f(a):=\left[\left(\frac{e^a}{W(e^{a+1})}\right)^{1/W(e^{a+1})} W(e^{a+1})\right]^p.$$

We will use that $$f(a)$$ is a convex function for all $$a\in\mathbb{R}$$ and $$p\ge 1$$, since its second derivative is non-negative. This can be verified in Mathematica by running

FullSimplify[D[((Exp[a]/ProductLog[Exp[a + 1]])^(1/ProductLog[Exp[a + 1]])*ProductLog[Exp[a + 1]])^p, {a, 2}], Assumptions -> {Element[a, Reals]}]


which gives $$f''(a) = \frac{p e^{p-\frac{p}{W\left(e^{a+1}\right)}} W\left(e^{a+1}\right)^{p-2} \left((p-1) W\left(e^{a+1}\right)+p\right)}{W\left(e^{a+1}\right)+1}$$ which is non-negative for $$p\ge 1$$.

We are now ready to prove the result: $$\sum_{i=1}^{n} \left(\frac{x_{i}^{x_{i+1}}}{x_{i+1}}\right)^p \ge n,\tag{2}$$ where we use the notation $$x_{n+1}=x_1$$. To begin, write $$\sum_i \left(\frac{x_{i}^{x_{i+1}}}{x_{i+1}}\right)^p \ge \sum_i f\left(\ln \frac{x_{i}}{x_{i+1}}\right) = n \frac{1}{n} \sum_i f\left(\ln \frac{x_{i}}{x_{i+1}}\right) \ge n f\left(\frac{1}{n} \sum_i \ln \frac{x_{i}}{x_{i+1}}\right)\tag{3}$$ where the first inequality uses $$(x^y/y)^p\ge f(\ln x/y)$$ while the second inequality is Jensen's. Note that $$\sum_i \ln \frac{x_{i}}{x_{i+1}}=0$$. It is also easy to verify from inspection that $$f(0)=1$$. Plugging into $$(3)$$ gives $$n f\left(\frac{1}{n} \sum_i \ln \frac{x_{i}}{x_{i+1}}\right)= n f(0) = n.$$ Combining with $$(3)$$ gives the desired result, $$(2)$$.

• Very nice. (+1) This is the 2nd time I saw this trick (something like $\min_{\alpha \in \mathbb{R}_+} (\alpha x)^{\alpha y}/(\alpha y)$). But I forgot where I saw this trick. Nov 1, 2021 at 6:28
• When proving $f$ convex, you cannot assume $a > 0$. Nov 1, 2021 at 6:38
• @AndersKaseorg Good point, actually the result does not depend on $a>0$ (I will update the answer) Nov 1, 2021 at 7:19

@Artemy's proof involves a complicated part of proving $$f''(a)\ge 0$$. Here is a proof which is a simplification of @Artemy's proof (one can do it by hand; also avoiding the use of the Lambert W function). Perhaps this is also helpful for the case when $$p < 1$$ (e.g. $$\sqrt{\frac{x^y}{y}}+\sqrt{\frac{y^x}{x}}\ge 2$$).

Fact 1: For any given $$x, y > 0$$, there exists a unique $$u > 0$$ such that $$\frac{x}{y} = u\mathrm{e}^{u - 1}$$. Furthermore, $$\frac{x^y}{y} \ge u\mathrm{e}^{1 - 1/u}.$$ (The proof is given at the end.)

Using Fact 1, let $$\frac{x_i}{x_{i + 1}} = u_i\mathrm{e}^{u_i - 1}, \,\, i = 1, 2, \cdots, n - 1; \qquad \frac{x_n}{x_1} = u_n\mathrm{e}^{u_n - 1}$$ where $$u_1, \cdots, u_n > 0$$. We have \begin{align*} 1 &= \frac{x_1}{x_2}\, \frac{x_2}{x_3} \cdots \frac{x_n}{x_1}\\ &= u_1 u_2 \cdots u_n \mathrm{e}^{u_1 + u_2 + \cdots + u_n - n}\\ &\le \left(\frac{u_1 + u_2 + \cdots + u_n}{n}\right)^n \mathrm{e}^{u_1 + u_2 + \cdots + u_n - n} \end{align*} which results in $$\frac{u_1 + u_2 + \cdots + u_n}{n} \ge 1. \tag{1}$$

Let $$f(u) = (u\mathrm{e}^{1 - 1/u})^p.$$ $$f(u)$$ is convex on $$u > 0$$ since $$f''(u) = (u\mathrm{e}^{1 - 1/u})^p pu^{-4}[(p - 1)u^2 + 2(p - 1)u + p] > 0$$ for all $$u > 0$$. Also, $$f(u)$$ is strictly increasing on $$u > 0$$ since $$f'(u) = (u\mathrm{e}^{1 - 1/u})^ppu^{-2}(u + 1) > 0$$ for all $$u > 0$$.

Using Fact 1 and (1), using the convexity and monotonicity of $$f(u)$$, we have \begin{align*} &\left(\frac{x_1^{x_2}}{x_2}\right)^p + \left(\frac{x_2^{x_3}}{x_3}\right)^p + \cdots + \left(\frac{x_n^{x_1}}{x_1}\right)^p\\ \ge\,& f(u_1) + f(u_2) + \cdots + f(u_n)\\ \ge\,& n\, f\left(\frac{u_1 + u_2 + \cdots + u_n}{n}\right) \\ \ge\,& n f(1)\\ \ge\,& n. \end{align*}

We are done.

Proof of Fact 1:

Since $$u\mapsto u\mathrm{e}^{u - 1}$$ is strictly increasing on $$u > 0$$, clearly, there exists a unique $$u > 0$$ such that $$\frac{x}{y} = u\mathrm{e}^{u - 1}$$.

We have $$\ln x = \ln y + \ln u + u - 1.$$ We need to prove that $$y\ln x - \ln y \ge \ln u + 1 - \frac{1}{u}.$$ It suffices to prove that $$y(\ln y + \ln u + u - 1) - \ln y \ge \ln u + 1 - \frac{1}{u}.$$

Let $$F(y) = y(\ln y + \ln u + u - 1) - \ln y - \left(\ln u + 1 - \frac{1}{u}\right).$$ We have $$F'(y) = \ln y + \ln u + u - \frac{1}{y},$$ and $$F''(y) = \frac{1}{y} + \frac{1}{y^2} > 0.$$ Also, $$F(1/u) = 0$$ and $$F'(1/u) = 0$$. Thus, $$F(y) \ge F(1/u) = 0$$ for all $$y > 0$$.

We are done.

• With respect to possible extensions to the case $p < 1$ I'll just mention that the inequality does not hold for $n=2$ and $p=1/4$, see my comment at the question. Nov 1, 2021 at 15:46
• @MartinR I know your comment. But how about $n = 2, 1/2 \le p < 1$, or $n \ge 3, p > 0$? Any counterexample? Nov 1, 2021 at 15:49
• I have no idea, I did not search for more (counter)examples. If I had to guess then I would say that $p \ge 1/2$ is sufficient. But I have no facts to back up that conjecture. Nov 1, 2021 at 15:52
• @MartinR Yes, before I think $p \ge 1/2$ might work. Actually, I even guess for $n \ge 3$, $p > 0$ works. A counterexample is expected. Nov 1, 2021 at 15:54
• @MartinR A thing I want to tell you: Anders Kaseorg's bound $\left(\frac{x^y}{y}\right)^p - 1 - p \ln x + p \ln y \ge 0$ does not hold for $p=1/2$. Nov 1, 2021 at 16:02

Partial answer Hint :

In the same vein as Anders Kaseorg we have :

Let $$a,x>1$$ or $$a\geq 1\geq x$$ or $$a\leq 1\leq x$$ or $$0.1\leq a\leq 1$$ and $$0 :

$$\sqrt{\frac{a^x}{x}} - \frac{a-1}{a+1} + \frac{x-1}{x+1}-1\geq 0$$

A sketch of proof can be found here Problem with an inequality : $\sqrt{\frac{a^x}{x}} - \frac{a-1}{a+1} + \frac{x-1}{x+1}-1\geq 0 \,?$

To be continued !

Edit 31/01/2022 :

Another case for $$0 and $$x>0$$ then we have :

$$f\left(x\right)=\sqrt{\frac{a^{x}}{x}}+\sqrt{\frac{x^{c}}{c}}-\frac{\left(a-1\right)}{a+1}+\frac{\left(c-1\right)}{c+1}-2\geq 0$$