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Let $a_j,b_j,c_j,d_j>0$, with $d_j\le c_j$ for $j=1,\cdots, n$.

Does $$\sum_{j=1}^k \frac{a_j}{b_j} \le\sum_{j=1}^k \frac{c_j}{d_j}, \quad k=1,\cdots, n$$ imply $$\frac{\sum_{j=1}^n a_j}{ \sum_{j=1}^n b_j} \le\frac{\sum_{j=1}^n c_j}{ \sum_{j=1}^n d_j}?$$

Editted My original question is

Does $$\prod_{j=1}^k \frac{a_j}{b_j} \le\prod_{j=1}^k \frac{c_j^2}{d_j^2}, \quad k=1,\cdots, n$$ imply $$\frac{\sum_{j=1}^n a_j}{ \sum_{j=1}^n b_j} \le \left(\frac{\sum_{j=1}^n c_j}{ \sum_{j=1}^n d_j}\right)^2?$$

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How is the first inequality related to the second? –  Listing Jul 30 '11 at 19:22
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3 Answers 3

up vote 3 down vote accepted

To your second question: This is also false:

$$\frac{6}{1} \cdot \frac{1}{1} \leq \frac{(5/8)^2}{(1/2)^2} \cdot \frac{2^2}{1^2}$$

but

$$\frac{6+1}{1+1} > (\frac{(5/8)+2}{(1/2)+1)})^2$$

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Note: I originally misread the constraint as $\sum_{i=1}^n \frac{a_i}{b_i} \leq \sum_{i=1}^n \frac{c_i}{d_i}$, while it actually is just one of the $n$ constraints. But it turns out that the conclusion $$\frac {\sum_i a_i}{\sum_i b_i} \leq \frac {\sum_i c_i}{\sum_i d_i}$$ again fails, and for pretty much the same reason. So here goes the modified solution...

One way of looking at the problem is as follows. Suppose $x_i = a_i/b_i$ and $y_j = c_j/d_j$. Then the problem stipulates that $\sum_{i=1}^k x_i \leq \sum_{j=1}^k y_j$ (mistake corrected). Now, we are asked whether it is necessarily the case that $$ \frac{\sum_i b_i x_i}{\sum_i b_i} \stackrel{?}{\leq} \frac{\sum_j d_j y_j}{\sum_j d_j}. $$

What does each side of the conjectured inequality mean? The LHS is the "weighted average" of the quantities $x_i$, according to the weights $b_i$; similarly for the RHS. It is easy to see the following proposition.

Suppose $x_i > 0$ are a set of numbers with minimum value $m$ and maximum value $M$. Then we have $m \leq \frac{\sum_i b_i x_i}{\sum_i b_i} \leq M$ assuming $b_i \geq 0$ and $\sum_i b_i > 0$. Moreover, for any $\mu \in [m, M]$, there exists some choice of $b_i \geq 0$ such that $\sum_i b_i = 1 > 0$ and $\frac{\sum_i b_i x_i}{\sum_i b_i} = \mu$.

Added condition: Suppose $m < \mu < M$ (that is, $m < M$ and these extreme points points are no longer allowed). Then there exists some choice of the weights $b_i > 0$ such that $\sum_{i} b_i = 1 > 0$ and $\frac{\sum_i b_i x_i}{\sum_i b_i} = \mu$. The point here is that none of the weights $b_i$ are allowed to be zero.

The proof of this proposition is simple, and I encourage you to work it out.

Now, we'll find a counterexample to the given problem. Assume that $x_i > 0$ is such that $\min_i x_i < \max_i x_i$ (that is, not all $x_i$s are equal). Let $y_i := x_i$ for all $i$, so that $\sum_{i=1}^k x_i \leq \sum_{j=1}^k y_j$ trivially holds for all $k$. Show that there exists some choice of weights $b_i > 0$ and $d_j > 0$ such that $\sum_{i} b_i = \sum_j d_j = 1$ and $$ \frac{\sum_i b_i x_i}{\sum_i b_i} > \frac{\sum_j d_j y_j}{\sum_j d_j} . $$

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No, because for example

$$(21/8)/1+3/1 \leq 2/(1/2)+2/1$$

but

$$(21/8+3)/(1+1) > (2+2)/(1/2+1)$$

Note that there are also many other counterexamples (I didn't look for the easiest one). For higher $n$ you can also easily see that it is wrong, the condition $d_j \leq c_j$ is confusing though.

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Thanks... I am trying to modify the problem such that it is correct.... –  Sunni Jul 30 '11 at 19:13
    
@Sunni: Good luck... –  Listing Jul 30 '11 at 19:15
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