Assuming that $\frac{a}{b}<\frac{c}{d} < 1 $, arrange $\frac{b}{a}$, $\frac{d}{c}$, $\frac{bd}{ac}$, $\frac{b+d}{a+c}$, and $1$

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Given that the positive integers $a,b,c,$ and $d$ satisfy $\cfrac{a}{b}<\cfrac{c}{d} < 1 $,arrange the following in order of increasing magnitude :$$\cfrac{b}{a},\cfrac{d}{c},\cfrac{bd}{ac},\cfrac{b+d}{a+c},1$$

My attempt

I've been able to get three of them in position ,namely $$\cfrac{bd}{ac} >\cfrac{b}{a}>\cfrac{d}{c}>1$$ since it's given that $\cfrac{a}{b} < \cfrac{c}{d}$ from which I have that $\cfrac{b}{a}>\cfrac{d}{c}$ and since $\cfrac{b}{a},\cfrac{d}{c} >1$ I have that $\cfrac{bd}{ac} >\cfrac{b}{a}>\cfrac{d}{c}>1$.

Now I am kind of clueless on how to tackle the expression $\cfrac{b+d}{a+c}$,as I don't see how I can get this by manipulating any of the given relations.

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From $$ b = a \frac ba > a \frac dc $$ it follows that $$ \frac{b+d}{a+c} > \frac{a \frac dc+d}{a+c} = \frac dc $$ and in a similar way you get $$ \frac{b+d}{a+c} < \frac ba \, . $$

There is also a general theorem which states that for positive numbers, $$ \min \frac{a_i}{b_i} \le \frac{a_1 + ... + a_n}{b_1 + ... + b_n} \le \max \frac{a_i}{b_i} $$ and this is proved using the same ideas.

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Just to continue from where you stopped $\frac{b+d}{a+c}$ is between $\frac dc$ and $\frac ba$. I guessed it using an example, and then proved the general result.

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$a,b,c$ can be any positive real numbers. They don't have to be positive integers.

What you wrote is correct. To finish the problem, notice $\frac{b}{a}>\frac{b+d}{a+c}>\frac{d}{c}$.

Proof: the first one is equivalent to $(a+c)b>(b+d)a$, i.e. $ab+bc>ab+ad$, i.e. $bc>ad$, i.e. $\frac{c}{d}>\frac{a}{b}$, which is true.

The second one is equivalent to $(b+d)c>(a+c)d$, i.e. $bc+cd>ad+cd$, i.e. $bc>ad$, i.e. $\frac{c}{d}>\frac{a}{b}$, which is true.

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If this is an exam question, the fastest way to do it is to choose a simple example: say, $\dfrac{a}{b}=\dfrac{1}{2}$ and $\dfrac{c}{d}=\dfrac{2}{3}$. Then you can just evaulate all those fractions and order them by size.