Homework Help: 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

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.