order statistics of random sample suppose $X_1,X_2,\ldots,X_n$ is a random sample of distribution with positive values where $E(X)=\text{Var}(X)=1$. We show order statistics of this random sample with $Y_1,Y_2,\ldots,Y_n$. How can show 


*

*$E\left(\displaystyle\sum_{i=1}^n \frac{Y_i}{X_i}\right)\geq n$

*$E\left(\displaystyle\sum_{i=1}^n Y_iX_i\right)\leq n+n^2$
 A: Let $S = X_1 + \ldots + X_n = Y_1 + \ldots + Y_n$.  We have $\text{Var}(S) = E[S] = n$.  Given $(Y_1,\ldots, Y_n)$, $(X_1,\ldots,X_n)$ is a random permutation of $(Y_1,\ldots,Y_n)$.
Then $E[Y_i X_i | Y_1,\ldots,Y_n] = Y_i (Y_1 + \ldots + Y_n)/n$ so $E[Y_i X_i ] = E[Y_i (Y_1 + \ldots + Y_n)/n] = E[Y_i S]/n$. Now
$$ E \left[ \sum_{i=1}^n Y_i X_i \right] = E \left[S^2/n \right]
= (\text{Var}(S) + E[S]^2)/n = (n + n^2)/n = 1 + n$$
EDIT: Similarly, $$E\left[\left. \sum_{i=1}^n \frac{Y_i}{X_i} \right| Y_1, \ldots,Y_n\right] =  \sum_{i=1}^n E\left[\left.\frac{Y_i}{X_i} \right| Y_1,\ldots,Y_n\right] = \frac{1}{n} \sum_{i=1}^n \sum_{j=1}^n \frac{Y_i}{Y_j}$$
Now note that $\displaystyle \sum_{i=1}^n \sum_{j=1}^n \frac{Y_i}{Y_j} = \sum_{j=1}^n \frac{X_i}{X_j}$, so
$$E \left [ \sum_{i=1}^n \frac{Y_i}{X_i} \right] = \frac{1}{n} E \left[\sum_{i=1}^n \sum_{j=1}^n \frac{Y_i}{Y_j}\right] = \frac{1}{n} \sum_{i=1}^n \sum_{j=1}^n E\left[\frac{X_i}{X_j}\right]$$
For the $n$ terms where $i = j$, $X_i/X_j = 1$.  For the other $n^2-n$ terms,  $X_i$ and $X_j$ are independent.  Thus
$$ E \left [ \sum_{i=1}^n \frac{Y_i}{X_i} \right] = 1 + (n-1) E[X] E[1/X]$$
By Jensen's inequality and the convexity of $1/x$, $E[1/X] \ge 1/E[X]$, so this gives you your inequality.
Moreover, this estimate is best possible, as can be seen by taking $$X = \cases{
1 + \sqrt{(1-p)/p} & with probability $p$\cr 1 - \sqrt{p/(1-p)} & with probability $1-p$\cr}$$
Then $E[X] = \text{Var}(X) = 1$, while $E[1/X] \to 1$ as $p \to 0+$.
A: The first one follows easily from the rearrangement inequality:
$${Y_1\over X_1}+\cdots+{Y_n\over X_n}\geq {Y_1\over Y_1}+\cdots+{Y_n\over Y_n}=n.\tag1$$
Now take expectations in (1). 
A: Here is direct way to establish the second inequality, although Robert's exact result implies it.
Since $X$ is a positive random variable, 
$$
    \sum_{k=1}^n Y_k X_k \leqslant \sum_{k=1}^n Y_k \sum_{i=1}^n X_i = S^2
$$
Taking the expectation:
$$
   \mathbb{E}\left(\sum_{k=1}^n Y_k X_k\right) \leqslant \mathbb{E}\left(S^2\right) = n^2+n
$$
where $\mathbb{E}\left(S^2\right) = \mathbb{Var}(S)+\mathbb{E}(S)^2 = n + n^2$ was used.
