# Is this equation provable?

1. Jul 26, 2011

### ArcanaNoir

1. The problem statement, all variables and given/known data
I'm working on a Cauchy inequality proof, and I'm at a point where I need to prove:
$a+ \frac{b}{n} = a\sqrt[n]{b}$ iff $a=b$

Anyway, is it even possible to prove this? If it isn't I need to hurry up and get about another method. If it is possible, I don't mind thinking on it some more.

Last edited: Jul 26, 2011
2. Jul 26, 2011

### micromass

Staff Emeritus
Hi ArcanaNoir!

This doesn't seem true. Take a=b=n=2, then

$$a+\frac{b}{n}=3$$

but

$$a\sqrt{n}{b}=2\sqrt{2}$$

and these are certainly not equal. So at least one of the implications fails.

Looking at the graph of

$$x+\frac{y}{2}-x\sqrt{y}$$

suggest that the other implication fails as well.

3. Jul 27, 2011

### ArcanaNoir

Oh dear.

4. Jul 27, 2011

### ArcanaNoir

Okay, I think I fixed it.
$$\frac{na+b}{n+1}=\sqrt[n+1]{a^nb}$$

5. Jul 27, 2011

### Clever-Name

I'm not well versed in proof writing but couldn't you just algebraically show it's true for a=b?

As in:

$$\frac{na + a}{n+1} = \sqrt[n+1]{a^na}$$
$$\frac{(n+1)a}{n+1} = \sqrt[n+1]{a^{n+1}}$$
$$a = a$$

Though this would require that n cannot equal -1, but I think you may already have that restriction.

6. Jul 27, 2011

### BrianMath

This would only prove one implication. This is an if and only if proof, so you have to prove that
$$a=b \implies \frac{na + b}{n+1}=\sqrt[n+1]{a^nb}$$
AND
$$\frac{na + b}{n+1} = \sqrt[n+1]{a^nb} \implies a=b$$

Last edited: Jul 27, 2011
7. Jul 27, 2011

### Clever-Name

isn't that what the op wants though? This statement is true iff a=b...

8. Jul 27, 2011

### ArcanaNoir

BrianMath is right. It's the second thing I need most, the first part of the biconditional is easy.

A and b are greater than zero and n is a natural number.