# Thread: Inequality Proof (not using Calculus)

1. ## Inequality Proof (not using Calculus)

Hey guys,

Just wondering would you be able to help me with this question. I know of a proof using calculus, but I think there may be another method.

$\left( {\frac{a+1}{b+1}}\right) ^{b+1} > \left( \frac{a}{b} \right) ^{b}$

for $a > b > 0$

2. Anyone?

3. Sorry, my last proof was faulty. I'll have a look into it

4. Originally Posted by alan4cult
Hey guys,

Just wondering would you be able to help me with this question. I know of a proof using calculus, but I think there may be another method.

$\left( {\frac{a+1}{b+1}}\right) ^{b+1} > \left( \frac{a}{b} \right) ^{b}$

for $a > b > 0$
(Edited) Hmm, I thought I had come up with a nifty proof, but seems now that I wasn't able to avoid calculus.

$\left( {\frac{a+1}{b+1}}\right) ^{b+1} > \left( \frac{a}{b} \right) ^{b}$

$\iff {\frac{(a+1)^{b+1}}{(b+1)^{b+1}}}> \frac{a^b}{b^b}$

$\iff (a+1)^{b+1}b^b > (b+1)^{b+1}a^b$

$\iff \frac{(a+1)^{b+1}}{a^b} > \frac{(b+1)^{b+1}}{b^b}$

This is of the form

$f(a) > f(b)$

where

$f(x) = \frac{(x+1)^{b+1}}{x^b}$

Now all we need to establish is that

$a > b \iff f(a) > f(b)$

In other words, we need to show that $f(x)$ is increasing.

"Clearly, the numerator of $f(x)$ grows faster than the denominator, thus $f(x)$ is increasing and the proof is concluded."

But I don't think that's quite good enough, and now I'm not sure how to prove rigorously that $f(x)$ is increasing without derivatives... maybe someone else can see a way?

EDIT: I'm not confident about the following method, but perhaps it is valid:

Let $g(u) = \log_x{u}$.

Since $g(u)$ preserves order (is monotonic), applying $g(u)$ to both sides of an equation will also preserve order. So:

$f(x) = \frac{(x+1)^{b+1}}{x^b}$

$\iff g(f(x)) = g\left(\left(\frac{x+1}{x}\right)^b (x+1)\right)$

$= (b+1)\log_x(x+1)-b\log_x{x}$

$= (b+1)\log_x(x+1)-b$

Let $h(x)=\log_x(x+1)$. Then it is sufficient to show that $h(x)$ is increasing, in order to prove that $f(x)$ is increasing.

Now let $k(u) = x^u$. This is again order preserving, and we have:

$h(x)=\log_x(x+1)$

$\iff k(h(x))=k\left(\log_x(x+1)\right)$

$=x+1$

It is clear that $x+1$ is increasing, therefore $f(x)$ is increasing and the proof is concluded.

I'm bothered by the result because, when I took derivatives, I found that $f'(x)=0$ when $x=b$, and nowhere in the above steps did I utilize the restriction $x \geq b$.

Ah well, I've edited this post about a million times, I'll leave it in the hands of other, smarter people.