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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 $1\le k<n$ 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.

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  • $\begingroup$ 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. $\endgroup$ Commented Oct 16, 2021 at 10:38
  • $\begingroup$ @TheSimpliFire It looks good. Why someone downvoted it? $\endgroup$
    – River Li
    Commented Oct 16, 2021 at 14:33
  • $\begingroup$ 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 $p \ge 1$. $\endgroup$
    – Martin R
    Commented Oct 18, 2021 at 11:56
  • $\begingroup$ @MartinR What if $n\ge 3$? $\endgroup$
    – River Li
    Commented Oct 18, 2021 at 12:06
  • $\begingroup$ @RiverLi: I only played a bit with the case $n=2$ so far. It could be that $p \ge 0.5$ is sufficient. $\endgroup$
    – Martin R
    Commented Oct 18, 2021 at 12:08

4 Answers 4

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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

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

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    $\begingroup$ 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. $\endgroup$
    – TheSimpliFire
    Commented Nov 2, 2021 at 11:17
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    $\begingroup$ @Hans I am using $u^p - 1 ≥ p(u - 1)$ to observe that the result for $p = 1$ implies the result for $p ≥ 1$. $\endgroup$ Commented Nov 3, 2021 at 0:32
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    $\begingroup$ I don't know why 100 reputation is given to my answer. It should be given to Anders Kaseorg's excellent answer. $\endgroup$
    – River Li
    Commented Nov 3, 2021 at 16:54
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    $\begingroup$ @RiverLi I agree; I suspect ErikSatie wanted to balance out the reputation. But thanks for starting the bounty, this answer totally deserves it. $\endgroup$
    – TheSimpliFire
    Commented Nov 3, 2021 at 17:21
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    $\begingroup$ 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? $\endgroup$
    – Hans
    Commented Nov 4, 2021 at 0:01
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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, $(2)$.

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    $\begingroup$ 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. $\endgroup$
    – River Li
    Commented Nov 1, 2021 at 6:28
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    $\begingroup$ When proving $f$ convex, you cannot assume $a > 0$. $\endgroup$ Commented Nov 1, 2021 at 6:38
  • $\begingroup$ @AndersKaseorg Good point, actually the result does not depend on $a>0$ (I will update the answer) $\endgroup$
    – Artemy
    Commented Nov 1, 2021 at 7:19
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@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.

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  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Commented Nov 1, 2021 at 15:46
  • $\begingroup$ @MartinR I know your comment. But how about $n = 2, 1/2 \le p < 1$, or $n \ge 3, p > 0$? Any counterexample? $\endgroup$
    – River Li
    Commented Nov 1, 2021 at 15:49
  • $\begingroup$ 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. $\endgroup$
    – Martin R
    Commented Nov 1, 2021 at 15:52
  • $\begingroup$ @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. $\endgroup$
    – River Li
    Commented Nov 1, 2021 at 15:54
  • $\begingroup$ @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 $p=1/2$. $\endgroup$
    – River Li
    Commented Nov 1, 2021 at 16:02
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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 $0<x\leq 1$ :

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

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 $0<c\leq a \leq 1$ and $x>0$ then we have :

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

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