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?



Monotonicity of complete homogeneous symmetric polynomials
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](http://qchu.wordpress.com/2009/11/20/the-jacobi-trudi-identities/), 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




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A few comments: [This paper](http://www.jstor.org/pss/2153964) 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. 


