Proving the Cauchy Inequality for Natural Numbers

[ArcanaNoir]

Homework Statement

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.

 

[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.

 

[ArcanaNoir]

Oh dear.

 

[ArcanaNoir]

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

 

[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.

 

[BrianMath]

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.

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
$$\frac{na + b}{n+1} = \sqrt[n+1]{a^nb} \implies a=b$$

 

[Clever-Name]

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

 

[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.