# Proving a desired inequality

1. Jul 6, 2008

### nasshi

I'm interested in proving the following inequality, given the following constraints. This is for personal gain, and not for homework. I'm not even sure if I can prove this without additional constraints. If you can solve it by adding constraints, please let me know what options I have for adding constraints.

Given:
0<a<1
x>0, y>0, z>0
x>y

Prove:$$\frac{x-y}{(1+x)(1+y)} < \frac{z*a}{1+y+z}$$

Thanks so much for your help!

Last edited: Jul 6, 2008
2. Jul 6, 2008

### epkid08

I believe it's proven wrong with the example, (a,x,y,z)=(1/2,2,1,1).

3. Jul 6, 2008

### nasshi

You are indeed correct.

$$y+z>x$$

4. Jul 7, 2008

### epkid08

Well y+z>x really just means z>x in your case, but I'll try to prove it:

$$\frac {x-y}{(1+x)(1+y)}<\frac {z*a}{1+y+z}$$

Fallowing constraints, we set (a,x,y,z) to (1/2,2,1,2)

$$\frac {2-1}{(1+2)(1+1)}<\frac {2*1/2}{1+1+2}$$

$$\frac {1}{6}<\frac {1}{4}$$

This proves true.
Assume $$(a,x,y,z)\longrightarrow(a+1,x+1,y+1,z+1)$$; plug in the original set.

$$\frac {(x+1)-(y+1)}{(1+(x+1))(1+(y+1))}<\frac {(z+1)(a+1)}{1+(y+1)+(z+1)}$$

$$\frac {x-y}{(x+2)(y+2)}<\frac {(z+1)(a+1)}{y+z+3}$$

Now plug in, and simplify:

$$\frac {2-1}{(2+2)(1+2)}<\frac {(2+1)(1/2+1)}{1+2+3}$$

$$\frac {1}{12}<\frac {9}{12}$$

I don't know if this is formal, but I'd call it adequate.

5. Jul 7, 2008

### nasshi

I think it works well for larger values of x,y,z. But when x,y,z are small, for example, between 0 and 1, I think your method won't work.

6. Jul 7, 2008

### epkid08

I should have set (a,x,y,z) to (d,2d,d,3d) and (1-d,2d,d,3d).
Anyways, you have to fix your constraints...

7. Jul 7, 2008

### nasshi

I can't exactly "fix my constraints". I'm trying to prove something as generally as possible. Can you think of any constraints I can add that would easily establish the desired result? Because I can't.

8. Jul 7, 2008

### epkid08

Here are some constraints that I think work:
x>y
0<y<1
a>6
z>2/35