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If $0<a<b$ and $0<c<d$ then $\frac{c+a}{d+a} <\frac{c+b}{d+b}.$
I get to $$d+a<d+b \Longrightarrow \frac{1}{d+b} < \frac{1}{d+a}$$ but that inequality seems opposite of what I am trying to prove. Any advice is appreciated.
 A: $\renewcommand{\iff}{\Leftrightarrow}$Here's a dumb way which requires no clever insight:
Since $a, b, c, d > 0$, we see that
$$\frac{c+a}{d+a} < \frac{c+b}{d+b} \iff(c + a)(d + b) < (d + a)(c + b).$$
Thus, it suffices to prove that the right side is true. Multiplying the brackets and cancelling the common terms, this becomes equivalent to showing that
$$cb + ad < db + ac.$$
Rearranging shows that the above is equivalent to
$$ad - ac < db - cb.$$
Note
$$ad - ac < db - cb \iff a(d - c) < b(d - c) \iff 0 < (b - a)(d - c).$$
The rightmost statement is true because $b - a > 0$ and $d - c > 0$.
A: $$ \dfrac{c+a}{d+a} =\dfrac{d+a+c-d}{d+a}  = 1 + \dfrac{c-d}{d+a}$$
$$ \dfrac{c+b}{d+b} =\dfrac{d+b+c-d}{d+b}  = 1 + \dfrac{c-d}{d+b}$$
You have already proved that
$$ \frac{1}{d+b} < \frac{1}{d+a}$$
As $c-d<0$,
$$ \frac{c-d}{d+b} > \frac{c-d}{d+a}$$
$$ \implies1+ \frac{c-d}{d+b} > 1+\frac{c-d}{d+a}$$
$$ \implies \dfrac{c+b}{d+b} > \dfrac{c+a}{d+a}$$
A: There is this important general inequality (working for positive denominators), i.e. the compound fraction comes between the bounds:
$$\frac AB<\frac CD\implies \frac AB<\frac{A+C}{B+D}<\frac CD$$
See here for a proof https://math.stackexchange.com/a/3373549/399263, but it is basically using the fact that $AD-BC<0$ once reduced to common denominator.
You can use it here by following this chain:
$$\frac cd<1=\frac aa\implies \frac {c+a}{d+a}<1=\frac{b-a}{b-a}\implies \frac{c+a}{d+a}<\frac{c+a+b-a}{d+a+b-a}=\frac{c+b}{d+b}$$
A: By the rearrangement inequality, $bc+ad\lt ac+bd.$ Thus, $(c+a)(d+b)\lt(c+b)(d+a).$
A: It may be helpful to just rename some of the quantities.
Suppose
$$\alpha = c+a\\
\beta=d+a\\
\gamma=b-a$$
then
$$\begin{align}
\frac{\alpha}{\beta}&\lt\frac{\alpha+\gamma}{\beta+\gamma}\\
\alpha\beta + \alpha\gamma&\lt \alpha\beta +\beta\gamma\\
\alpha\gamma&\lt\beta\gamma\\
\alpha&\lt\beta\\
\end{align}$$
Now, reversing these steps should give you a road map to writing a formal proof.
