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Question

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.

Answer 1

$\renewcommand{\iff}{\Leftrightarrow}$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 $d - c > 0$.

Answer 2

$$ \dfrac{c+a}{d+a} =\dfrac{d+a+c-d}{d+a} = 1 + \dfrac{c-d}{d+a}$$ $$ \dfrac{c+b}{d+b} =\dfrac{d+b+c-d}{d+b} = 1 + \dfrac{c-d}{d+b}$$ You have already proved that $$ \frac{1}{d+b} < \frac{1}{d+a}$$ As $c-d<0$, $$ \frac{c-d}{d+b} > \frac{c-d}{d+a}$$ $$ \implies1+ \frac{c-d}{d+b} > 1+\frac{c-d}{d+a}$$ $$ \implies \dfrac{c+b}{d+b} > \dfrac{c+a}{d+a}$$

Answer 3

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:

$$\frac cd<1=\frac aa\implies \frac {c+a}{d+a}<1=\frac{b-a}{b-a}\implies \frac{c+a}{d+a}<\frac{c+a+b-a}{d+a+b-a}=\frac{c+b}{d+b}$$

Answer 4

By the rearrangement inequality, $bc+ad\lt ac+bd.$ Thus, $(c+a)(d+b)\lt(c+b)(d+a).$

Answer 5

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.