  Let $G$ be a simple graph of order $n$. An independent set in a graph is a
set of pairwise non-adjacent vertices. The independence polynomial of $G$ is
the polynomial $I(G,x)=\sum_{k=0}^{n} s(G,k) x^{k}$, where $s(G,k)$ is the
number of independent sets of $G$ of size $k$ and $s(G,0)=1$. Clearly all real
roots of $I(G,x)$ are negative. Let $\xi(G)$ be the largest real root of
$I(G,x)$. Let $H$ be a simple graph. By $G \succeq H$ we mean that $I(H,x)\geq
I(G,x)$ for every $x$ in the interval $[\xi(G),0]$. We note that $G\succeq H$
implies that $\xi(G)\geq \xi(H)$. Also we let $G\succ H$ if and only if
$G\succeq H$ and $I(G,x)\neq I(H,x)$. We prove that for every tree $T$ of order
$n$, $S_n\succeq T\succeq P_n$, where $S_n$ and $P_n$ are the star and the path
of order n, respectively. By $T=T(n_1,\ldots,n_k)$ we mean a tree $T$ which has
a vertex $v$ of degree $k$ such that $T\setminus v=P_{n_1-1}+\cdots+P_{n_k-1}$,
that is $T\setminus v$ is the disjoint union of the paths
$P_{n_1-1},\ldots,P_{n_k-1}$. Let $X=(x_1,\ldots,x_k)$ and
$Y=(y_1,\ldots,y_k)$, where $x_1\geq \cdots\geq x_k$ and $y_1\geq\cdots\geq
y_k$ are real. By $X\succ Y$, we mean $x_1=y_1,\ldots,x_{t-1}=y_{t-1}$ and
$x_t>y_t$ for some $t\in\{1,\ldots,k\}$. We let $X\succ_{d}Y$, if $X\neq Y$ and
for every $j$, $1\leq j\leq k$, $\sum_{i=1}^{j}x_i\geq \sum_{i=1}^{j}y_i$.
Among all trees with fixed number of vertices, we show that if
$(m_1,\ldots,m_k)\succ_{d}(n_1,\ldots,n_k)$, then $T(n_1,\ldots,n_k)\succ
T(m_1,\ldots,m_k)$. We conjecture that $T(n_1,\ldots,n_k)\succ
T(m_1,\ldots,m_k)$ if and only if $(m_1,\ldots,m_k)\succ (n_1,\ldots,n_k)$,
where $\sum_{i=1}^kn_i=\sum_{i=1}^km_i$.
