Your statement is true. This inequality is equivalent to the fact that the correlation of $\frac{\sum S_{\geq p}}{\sum S}$ and $\sum S$ is positive correlated. More precisely:

For a given $q$, let $f(x) = 0$ for all $x<q$ and $f(x) = x$ for all $x\geq q$. Your inequality can be written as: \begin{equation} E\left(\frac{\sum f(X_i)}{\sum X_i}\right)\leq \frac{E f(X)}{E(X)} \end{equation} The left side equals to $nE\left(\frac{f(X_1)}{X_1+\sum_{i\geq 1} X_i}\right)$ since for $i=1,2,...$ the random variable $\frac{f(X_i)}{\sum X_i}$ has the same law. Since $x\mapsto f(x)/(x+y)$ and $x\mapsto x+y$ are increasing functions w.r.t $x$, the random variable $f(X_1)/(X_1+Y)$ and $X_1+Y$ (with $Y=\sum_{i\geq 2} X_i$, independent from $X_1$) is positive correlated. Then
$$ E\left(\frac{f(X_1)}{X_1+Y}(X_1+Y)\right) - E\left(\frac{f(X_1)}{X_1+Y}\right) E(X_1+Y)\leq 0.$$ Rearranging the terms gets your inequality.

Your statement is true. This inequality is equivalent to the fact that the correlation of $\frac{\sum S_{\geq p}}{\sum S}$ and $\sum S$ is positive correlated. More precisely:

For a given $q$, let $f(x) = 0$ for all $x<q$ and $f(x) = x$ for all $x\geq q$. Your inequality can be written as: \begin{equation} E\left(\frac{\sum f(X_i)}{\sum X_i}\right)\leq \frac{E f(X)}{E(X)} \end{equation} The left side equals to $nE\left(\frac{f(X_1)}{X_1+\sum_{i\geq 1} X_i}\right)$ since for $i=1,2,...$ the random variable $\frac{f(X_i)}{\sum X_i}$ has the same law. Since $x\mapsto f(x)/(x+y)$ and $x\mapsto x+y$ are increasing functions w.r.t $x$, the random variable $f(X_1)/(X_1+Y)$ and $X_1+Y$ (with $Y=\sum_{i\geq 2} X_i$, independent from $X_1$) is positive correlated. Then
$$ E\left(\frac{f(X_1)}{X_1+Y}(X_1+Y)\right) - E\left(\frac{f(X_1)}{X_1+Y}\right) E(X_1+Y)\geq 0.$$ Rearranging the terms gets your inequality.