Let $a_1, a_2, \ldots, a_n$ be positive real numbers. A while ago I noticed that if you form the polynomial $$ P(x) = (x - a_1)(x-a_2) \cdots (x-a_n) $$ then:

  • The arithmetic mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{n-1}$.
  • The geometric mean of $a_1, \ldots, a_n$ is the positive number $m$ such that $(x - m)^n$ and $P(x)$ have the same coefficient of $x^{0}.$

It looks like this can be extended: for any $0 \le i \le n-1$, let the $i$th mean be the number $m_i$ such that $(x - m_i)^n$ and $P(x)$ have the same coefficient of $x^i$. (Alternatively, one can define $m_i$ in terms of elementary symmetric polynomials.)

For instance, with three variables $x, y, z$ we get \begin{align*} m_0 &= \sqrt[3]{xyz} \\ m_1 &= \sqrt{\frac{xy + yz + zx}{3}} \\ m_2 &= \frac{x + y + z}{3} \end{align*}

I am not sure what qualifies something as a "mean", but $m_1$ is symmetric and lies strictly between the min and the max, and it probably has other properties as well. Here's a more concrete question:

Question: Must it be true that $m_0 \le m_1 \le m_2 \le \cdots \le m_{n-1}$?

  • 5
    $\begingroup$ It's a known Maclaurin's inequality. See my proof in "КванТ", 1980, 04, M565 $\endgroup$ – Michael Rozenberg Dec 26 '14 at 8:12
  • $\begingroup$ @MichaelRozenberg Fantastic! Thanks, now I know what this inequality is called, and that will probably be enough for me to find anything I need about it. Unfortunately I do not know how to access the text you referenced. $\endgroup$ – 6005 Dec 26 '14 at 8:18
  • $\begingroup$ @Goos:not sure if this would be of any help, but the referenced text can be found in the archive of the Kvant magazine 198004.djvu, see pages 35(34)--36(35). Beware that the text is in Russian, but you can follow the equations. $\endgroup$ – g.kov Dec 27 '14 at 2:40
  • $\begingroup$ en.wikipedia.org/wiki/Maclaurin%27s_inequality $\endgroup$ – 6005 Feb 1 '15 at 3:13
  • $\begingroup$ See this paper math.washington.edu/~morrow/papers/nate-thesis.pdf $\endgroup$ – Yuriy S Jan 27 '17 at 21:20

Denote, $(\overline{a}) = (a_1,\cdots,a_n)$ and $\displaystyle P(x) = \prod\limits_{k=1}^{n}(x - a_k) = x^n +\sum\limits_{k=1}^{n} (-1)^k\binom{n}{k}u_k(\overline{a})x^{n-k}$

where, $\displaystyle u_k(\overline{a}) = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n}a_{j_1}\cdots a_{j_k}}{\binom{n}{k}} = m_{n-k}^{k}$ (in your notation).

Note that $P(x)$ has $n$ real roots in the interval $\left[\min\limits_{i=1}^n\{a_i\},\max\limits_{i=1}^n\{a_i\}\right]$,

Thus $P^{(n-2)}(x) = \dfrac{n!}{2}(x^2 - 2u_1x + u_2)$ has two real roots in that interval, i.e., $u_1^2 \ge u_2$.

Similarly apply the same idea for the $(n-2)^{th}$ derivative of the polynomial with roots $\dfrac{1}{a_k}$, ($k = 1(1)n$)

We get, $\displaystyle \frac{1}{\binom{n}{2}}\sum\limits_{i<j}a_ia_j \le \frac{1}{\binom{n}{1}^2}\left(\sum\limits_{i=1}^{n} \frac{1}{a_i}\right)^2 \implies u_{n-1}^2 \ge u_nu_{n-2}$.

We show, $u_{k-1}(\overline{a})u_{k+1}(\overline{a}) \le u_k^2(\overline{a})$ for $k = 2,3,\cdots,n-1$,

We can prove this result by induction on $n$, suppose the inequality holds for any $n-1$ positive real numbers.

We have $\displaystyle P'(x) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}u_k(\overline{a})x^{n-k-1}\right)$

If the roots of $P'(x)$ are $b_k$, for $k=1,2,\cdots,n-1$, (which, are positive reals by M.V.T.).

and define $v_k = \dfrac{\sum\limits_{1\le j_1<\cdots< j_k \le n-1}b_{j_1}\cdots b_{j_k}}{\binom{n-1}{k}}$

Then $\displaystyle P'(x) = n\prod\limits_{k=1}^{n-1}(x - b_k) = n\left(x^{n-1} + \sum\limits_{k=1}^{n-1}(-1)^k\binom{n-1}{k}v_kx^{n-k-1}\right)$

Thus, $u_k = v_k$ for $k=1,2,\cdots,n-1$ and by the induction hypothesis on the numbers $(b_1,\cdots,b_{n-1})$ we get $u_k^2 \ge u_{k+1}u_{k-1}$ for $k = 2,\cdots,n-2$ and together with $u_{n-1}^2 \ge u_nu_{n-2}$ completes the induction.

If we take the $k$th inequality to the $k$th power and then multiply all these inequalities for $k=1,\cdots,r$, we get,

$u_1^2 u_2^4 u_3^6 \cdots u_{r-1}^{2r-2}u_{r}^{r-1}u_{r+1}^r \le u_1^2u_2^4\cdots u_r^{2r} \implies u_{r+1}^r \le u_{r}^{r+1}$ for $r=1, \cdots,n-1$

Thus, $u_r^{1/r} = m_{n-r}$ forms a non increasing sequence.

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