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Assume $0 < p_1 \le p_2\le p_3 \le p_4$. What is the maximum of the following expression?

$$ \frac{\left(p_1+p_4\right)\left(p_2+p_3\right)}{\left(p_1+p_3\right)\left(p_2+p_4\right)} $$

Is that bounded by a constant?

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1 Answer

up vote 3 down vote accepted

$$\frac{\left(p_1+p_4\right)\left(p_2+p_3\right)}{\left(p_1+p_3\right)\left(p_2+p_4\right)} = 1 + \frac{\left(p_2-p_1\right)}{\left(p_1+p_3\right)} \times \frac{\left(p_4-p_3\right)}{\left(p_2+p_4\right)} $$ and the right hand side is greater than or equal to $1 + 0 \times 0$ but less than $1 + 1 \times 1$.

The values $(1,n,n,n^2)$ achieve the lower bound when $n=1$ but approach the upper bound when $n$ increases without limit: $n=400$ gives a figure over $1.99$.

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Just for completeness: Thus, the answer to the first question is that the expression has no maximum. –  joriki Mar 13 '12 at 1:23
    
@joriki,@Henry: You mean, it has no maximum but it is upper bounded by 2, right? –  Mohsen Mar 13 '12 at 14:32
    
@Mohsen: Yes. In fact its supremum is $2$. By the way, you don't need to ping Henry when you comment under his answer; the author of a post is automatically notified of any comments under it. –  joriki Mar 13 '12 at 14:45
    
Thanks joriki, good point :) –  Mohsen Mar 13 '12 at 15:00
    
Thanks Henry for your answer, I asked a more general case of the above question here: math.stackexchange.com/questions/119702/… Could you please have a look at it. –  Mohsen Mar 13 '12 at 15:30
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