# Thread: Inequality proof.

1. ## Inequality proof.

I need to mathematically prove that:

(a + b)/(c + d) lies between (a / c) and (b / d).

a,b,c,d are all real and >0

Any ideas?

(also, if necessary you can assume that b>a and d>c but preferably not)

2. We don't need to assume $b\geq a$ (or the other way), but we do have to assume either $ad \leq bc$ or the opposite*. It's obviously one of them (unless it's both) so there's no danger in doing this; also notice that the roles of a and d can be switched with those of b and c respectively in the middle of the inequality, so what we're doing is all kosher.

So the question is: Prove $\frac{a}{c} \leq \frac{a+b}{c+d} \leq \frac{b}{d}$

Start with the assumption ad < bc (When I write < I mean less than or equal to, but I don't want to format this entire answer).

First, add ac to both sides:
ad+ac < bc+ac
a(d+c) < c(a+b)

a/c < (a+b)/(c+d). Voila, the left inequality

Then go back to the assumption but add bd to both sides:
ad+bd < bc+bd
d(a+b) < b(c+d)

(a+b)/(c+d) < b/d. And there's the right one.

[ Note: when I did my scratchwork, I did it by assuming what I wanted to prove and tried to get something I knew to be true; I didn't just pull "+bd" out of nowhere. But that's not actually a proof - you're not allowed to assume what you want to be true and then use it to show 0=0, the logic can fail ]

* This may seem like a bizarre thing to assume, but if I wrote it like this: "We must assume either a/c < b/d OR b/d < a/c," now it makes more sense, doesn't it? I chose the non-fraction form because it was easier for the computation in the proof, also it's easier to see how the a/b // d/c pairs can be switched around.