Some Inequalities for Power Means; a Problem from “The Logarithmic Mean Revisited”

Abstract

We establish some inequalities comparing power means of two numbers with combinations of the arithmetic and geometric means. A conjecture from [1] is confirmed.

MSC:

• 26D15

Given positive numbers a, b, the arithmetic, geometric and pth power means are$$A(a,b)=\frac{1}{2}(a+b),\hspace{1em}G(a,b)={(ab)}^{1/2},\hspace{1em}{M}_p(a,b)={(\frac{1}{2}(a^p+b^p))}^{1/p}$$for $p\ne 0$. With a, b fixed, we denote these just by A, G, and M_p.

Of course, these definitions extend in a natural way to more than two numbers. For any finite set of positive numbers, it is clear that $M_1=A$, and well known that $G\le M_p\le A$ for $0<p\le 1$, $M_p\ge A$ for $p\ge 1$ and $M_p\le G$ for p < 0. (Also, one defines M₀ to be G).

For two numbers, it is easily seen that $M_{1/2}=\frac{1}{2}G+\frac{1}{2}A$. This equality does not extend to more than two numbers: for the triple (4, 1, 1) we have $M_{1/2}<\frac{1}{2}G+\frac{1}{2}A$, while the opposite inequality holds for (4, 4, 1). From now on, we restrict considerations to two numbers a, b. It was shown in the note [1] that $M_{1/3}\le \frac{2}{3}G+\frac{1}{3}A$, and conjectured that $M_p\le (1-p)G+pA$ for $0<p\le \frac{1}{2}$, together with the opposite inequality for $\frac{1}{2}\le p\le 1$. As reported in [1], the conjecture was confirmed by Gord Sinnamon; his proof (communicated privately) is ingenious, but it involves some fairly heavy manipulation.

A more complete picture is obtained if at the same time we compare M_p with $G^{1-p}{A}^p$. Equality holds for p = 1, and it is easily verified that $M_{-1}=G^2/A$ (this is the harmonic mean), so equality also holds for p = – 1. The results in [1] imply that $M_{1/3}\ge G^{2/3}{A}^{1/3}$ (with the logarithmic mean coming between these two quantities), suggesting that a similar inequality holds for $0<p\le 1$, though this was not explicitly stated as a conjecture.

Here we offer a simple unified treatment of both comparisons, based on the substitution that was used in [1]. The results are as follows.

Theorem 1.

The inequality $M_p\le (1-p)G+pA$ holds for $0<p\le \frac{1}{2}$ and for $p\ge 1$. The opposite inequality holds for $\frac{1}{2}\le p\le 1$ and for p < 0.

Theorem 2.

The inequality $G^{1-p}{A}^p\le M_p$ holds for $0<p\le 1$ and for $p\le -1$. The opposite inequality holds for $p\ge 1$ and for $-1\le p<0$.

Note first that if $x=a/b$, then $A(a,b)=bA(x,1)$ and similarly for G and M_p, so it is sufficient to consider the pair $(x,1)$: henceforth the notation A, G, M_p applies to this pair. Now substitute $x=e^{2t}$. Then $G=e^t$ and$$A=\frac{1}{2}(e^{2t}+1)=e^t\cosh t,\mathrm{\hspace{1em}}\mathrm{\hspace{1em}}{M}_p={(\frac{1}{2}(e^{2pt}+1))}^{1/p}=e^t{(\cosh pt)}^{1/p}.$$

So, for example, the inequality $M_p\ge G$ stated above translates to ${(\cosh pt)}^{1/p}\ge 1$, which is obvious for p > 0. The inequality in Theorem 1 translates to(1) $${(\cosh pt)}^{1/p}\le (1-p)+p\cosh t.$$(1)

For both theorems, we will use the following Lemma, essentially an adaption of the “integrating factor” method to inequalities.

Lemma 3.

Let f be a function satisfying $f(0)=f\prime (0)=0$ and $f″(t)\ge f(t)$ for $t\ge 0$. Then $f(t)\ge 0$ for t > 0. The reverse applies if $f″(t)\le f(t)$ for t > 0.

Proof.

Let $g(t)=f\prime (t)+f(t)$ and $h(t)=f\prime (t)-f(t)$. Then $g(0)=h(0)=0$ and$$g\prime (t)-g(t)=h\prime (t)+h(t)=f″(t)-f(t)\ge 0,$$hence$$\begin{array}{c}\frac{d}{dt}(e^{-t}g(t))=e^{-t}(g\prime (t)-g(t))\ge 0,\\ \frac{d}{dt}(e^{t}h(t))=e^t(h\prime (t)+h(t))\ge 0.\end{array}$$

Consequently $e^{-t}g(t)$ and $e^{t}h(t)$ are increasing. So for t > 0, we have $g(t)\ge 0$ and $h(t)\ge 0$, hence $f\prime (t)\ge 0$, so also $f(t)\ge 0$. The inequalities reverse if $f″(t)\le f(t)$. □

Proof of Theorem 1.

As we have seen, the substitution $x=e^{2t}$ translates $M_p\le$ $(1-p)G+pA$ to $f(t)\ge 0$ (for all t), where$$f(t)=p\cosh t+(1-p)-{(\cosh pt)}^{1/p}.$$

Since f is even, it is enough to consider t > 0. Then $f(0)=0$ and$$f\prime (t)=p\sinh t-{(\cosh pt)}^{1/p-1}\sinh pt.$$

So $f\prime (0)=0$ and$$\begin{array}{c}f″(t)=p\cosh t-p{(\cosh pt)}^{1/p}-(1-p){(\cosh pt)}^{1/p-2}{(\sinh pt)}^2\\ =p\cosh t-{(\cosh pt)}^{1/p}+(1-p){(\cosh pt)}^{1/p-2}\\ =f(t)-(1-p)+(1-p){(\cosh pt)}^{1/p-2}.\end{array}$$

If $0<p\le \frac{1}{2}$, then $\frac{1}{p}-2\ge 0$, so ${(\cosh pt)}^{1/p-2}\ge 1$ and $f″(t)\ge f(t)$ for all t. If $p\ge \frac{1}{2}$ or p < 0, then ${(\cosh pt)}^{1/p-2}\le 1$. So if $\frac{1}{2}\le p\le 1$ or p < 0, then $f″(t)\le f(t)$, and if $p\ge 1$, then $f″(t)\ge f(t)$. The statements follow, by the Lemma. □

Proof of Theorem 2.

The inequality $G^{1-p}{A}^p\le M_p$ translates to(2) $${(\cosh pt)}^{1/p}\ge {(\cosh t)}^p$$(2)

(this inequality is perhaps of some interest in its own right). Let$$f(t)={(\cosh pt)}^{1/p^2}-\cosh t.$$

Then $f(0)=0$ and$$\begin{array}{c}f\prime (t)=\frac{1}{p}{(\cosh pt)}^{1/p^2-1}\sinh pt-\sinh t,\\ f″(t)={(\cosh pt)}^{1/p^2}+r(t)-\cosh t=f(t)+r(t),\end{array}$$where$$r(t)=(\frac{1}{p^2}-1){(\cosh pt)}^{1/p^2-2}{(\sinh pt)}^2.$$

If $\left|p\right|\le 1$, then $1/p^2-1\ge 0$, so $r(t)\ge 0$, hence $f″(t)\ge f(t)$, so $f(t)\ge 0$ for $t\ge 0$. This implies (2) if $0<p\le 1$ and the reverse of (2) if $-1\le p<0$. If $\left|p\right|\ge 1$, then $f″(t)\le f(t)$, so $f(t)\le 0$ for $t\ge 0$. This implies the reverse of (2) for $p\ge 1$ and (2) for $p\le -1$. □

It remains to compare and combine the inequalities in Theorems 1 and 2. There are five intervals to consider. For $0<p\le \frac{1}{2}$, we have$$G^{1-p}{A}^p\le M_p\le (1-p)G+pA.$$

For $-1\le p<0$, we have$$(1-p)G+pA\le M_p\le G^{1-p}{A}^p.$$

In the other cases, we have either two upper bounds or two lower ones. We compare them. For this purpose, write $(1-p)G+pA=C$. For $\frac{1}{2}\le p\le 1$, we have $G^{1-p}{A}^p\le C$, so the better estimate is $(1-p)G+pA\le M_p$, given by Theorem 1. (Recall that in this case we have the upper bound $M_p\le A$).

For $p\ge 1$, we have $C\le G^{1-p}{A}^p$, by the weighted AM-GM inequality applied to $A=\frac{1}{p}C+(1-\frac{1}{p})G$. So the better estimate is $M_p\le pA-(p-1)G$, again from Theorem 1. (Also $M_p\ge A$).

For $p\le -1$, we have again $C\le G^{1-p}{A}^p$, seen by writing $G=[1/(1+q)]C+[q/(1+q)]A$, where $q=-p$. So the better estimate is $G^{1-p}{A}^p\le M_p$, given by Theorem 2. (Also $M_p\le G$).