The complete homogeneous symmetric polynomials are defined as $$ h_k (x_1, \dots,x_n) = \sum_{1 \leq i_1 \leq i_2 \leq \cdots \leq i_k \leq n} x_{i_1} x_{i_2} \cdots x_{i_k}. $$ For example, $$ h_3(x_1,x_2,x_3) = x_1^3 + x_2^3 + x_3^3 + x_1^2 x_2 + x_1^2 x_3 + x_2^2 x_3 + x_2^2 x_1 + x_3^2 x_1 + x_3^2 x_2 + x_1 x_2 x_3. $$ I'm interested in conditions as simple as possible (and ideally independent of $n$) that will suffice to prove that $$ h_1(x_1,\dots,x_n) > h_2(x_1,\dots,x_n) > h_3(x_1,\dots,x_n) > \cdots $$ for some given set of positive real numbers $\{ x_1, \dots, x_n \}$.

For example, do these inequalities hold if $h_1(x_1,\dots,x_n) \le 1$? Do they all hold if the first one $h_1(x_1,\dots,x_n) \ge h_2(x_1,\dots,x_n)$ holds?


I claim that, if $h_k > h_{k+1}$, then $h_j > h_{j+1}$ for all $j \geq k$. This includes both your assertions above. (Note that $h_0$ should be considered to be $1$.)

More precisely, I will prove that, for any positive reals $x_j$, we have $$\frac{h_0}{h_1} \leq \frac{h_1}{h_2} \leq \frac{h_2}{h_3} \leq \cdots $$ So, if any term is greater than $1$, so all all the following terms.

Proof: The inequality $h_{k-1}/h_k \leq h_k/h_{k+1}$ is equivalent to $\det \left( \begin{smallmatrix} h_{k} & h_{k-1} \\ h_{k+1} & h_k \end{smallmatrix} \right) \geq 0$. By the Jacobi-Trudi identity, this determinant is equal to the Schur function $s_{k,k}$. In particular, since all the monomials in $s_{k,k}$ have nonnegative coefficients, the determinant is nonnegative as desired. QED

A few comments: This paper by Bruce Sagan looks relevant, although I don't have JSTOR access right now to check.

As a general heuristic, when presented with an inhomogenous inequality between symmetric functions, it is a good idea to try to think what homogenous inequality it might be a special case of. So trying to show "$1 \geq h_1$ implies $h_k \geq h_{k+1}$" suggests instead showing "$h_1 h_k \geq h_{k+1}$", since the latter is homogenous.

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    $\begingroup$ Beautiful. Simply beautiful. $\endgroup$ – Greg Martin Jun 2 '11 at 17:00
  • $\begingroup$ @David Speyer: Just thought I'd let you know that Bill Banks and I have just submitted our manuscript in which this result was strongly used. Thank you again for your response! - soon to appear ont he arXiv; also at math.ubc.ca/~gerg/index.shtml?abstract=OPSRP $\endgroup$ – Greg Martin Jan 6 '13 at 22:29

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