# Thread: Fraction proof

1. ## Fraction proof

I don't know, it has variables in it... Ugh finding a section is hard.

Prove $\displaystyle \frac{x+ab}{y+ab} \geq \frac{x}{y}$

because I moved countries a lot when I was little I think I missed out on most of the fraction axioms taught in elementary (took me to the end of high school to figure out cross multiplication >.>) so it might be that I'm missing some fundamental rule here...

2. LEt x=1, y=2,a=1, b=-1

So $\displaystyle \frac{1-1}{2-1}\geq \frac{1}{2}$ is false,

so what are the restrictions on x,y,a,b, are they all positive, are just a,b>0... we need more info otherwise I just gave you a counterexample meaning its false

3. Originally Posted by artvandalay11
LEt x=1, y=2,a=1, b=-1

So $\displaystyle \frac{1-1}{2-1}\geq \frac{1}{2}$ is false,

so what are the restrictions on x,y,a,b, are they all positive, are just a,b>0... we need more info otherwise I just gave you a counterexample meaning its false
oh, yeah, a,b > 0

That was important, wasn't it >.>

4. I'm thinking it's a proof by induction but I don't even know where to go with that >.> I mean, what am I adding 1 to? There are 4 variables here! And cases means too many cases...

5. i dont think it's proof by induction, although we could just let c=ab for some c>0 but that is still 3 variables

Let's go with this

Consider $\displaystyle y(x+ab)-x(y+ab)$

$\displaystyle =yx+yab-xy-xab=yab-xab=ab(y-x)$

If $\displaystyle ab(y-x)=0$ then $\displaystyle y=x$, so $\displaystyle \frac{x}{y}=1$ and $\displaystyle \frac{x+ab}{y+ab}=1$ so the inequality holds

If $\displaystyle ab(y-x)>0$, then $\displaystyle y(x+ab)-x(y+ab)>0$ so $\displaystyle y(x+ab)>x(y+ab)$ and so $\displaystyle (x+ab)>\frac{x(y+ab)}{y}$

And so $\displaystyle \frac{x+ab}{y+ab}>\frac{x}{y}$ (note that everything we divided by was >0 so that argument really is valid)

If ab(y-x)<0 then x>y since ab>0. So $\displaystyle y(x+ab)-x(y+ab)<0$ and just looking ahead this fails your inequality meaning we must place the restriction that x cannot be >y

For example let x=4 a=1 b=1 and y=1

Then $\displaystyle \frac{4+1(1)}{1+1(1)}=\frac{5}{2}=2.5$

$\displaystyle \frac{x}{y}=\frac{4}{1}=4$ and 2.5 is not $\displaystyle \geq 4$

6. That's supposed to be a greater than in the second part, yeah?

7. yes sorry about that very important typo