Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question

1 Answer 1

up vote 10 down vote accepted

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.

share|cite|improve this answer
Beautiful. Simply beautiful. – Greg Martin Jun 2 '11 at 17:00
@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 – Greg Martin Jan 6 '13 at 22:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.