Inequality with real exponents I had the following problem on a past homework asignment:
For $x,p \in \mathbb{R}^+$
$$\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
So, I think one way to show this for $1\leq p$ would be with rules for real exponents for real exponents (since $x^{-1} >-1$):
$$(1+\frac{1}{x})^p \geq 1+p \frac{1}{x}>1+\frac{p}{1+x}$$
and since $(1+x)^p>0$
$$\frac{(1+x)^{p}}{x^p}=\left(\frac{1+x}{x}\right)^p=(1+x^{-1})^p>1+\frac{p}{1+x}$$
$$\iff\frac{1}{x^p}>\frac{1}{(1+x)^{p}}+\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{p}{(1+x)^{p+1}}<\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
For $0<p<1$ this argument doesn't hold since the inequality gets flipped:
$$(1+\frac{1}{x})^p \leq 1+p \frac{1}{x}>1+\frac{p}{1+x}$$
$$\frac{(1+x)^{p}}{x^p}=\left(\frac{1+x}{x}\right)^p=(1+x^{-1})^p\leq 1+px^{-1}>1+\frac{p}{1+x}$$
$$\iff\frac{1}{x^p}\leq\frac{1}{(1+x)^p}+\frac{p}{x(1+x)^p}>\frac{1}{(1+x)^p}+\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{1}{x^p}-\frac{1}{(1+x)^p}\leq\frac{p}{x(1+x)^p}>\frac{p}{(1+x)^{p+1}}$$
$$\iff\frac{p}{(1+x)^{p+1}}>\frac{1}{x^p}-\frac{1}{(1+x)^{p}}$$
What would I need to do to complete this proof? I tried looking at $x^p(p+x+1)<(x+1)^{p+1}$ as well and got nowhere. Is this 'possible' to prove without the binomial series?

Let $f(x)=\dfrac{1}{x^p}$, then $f'(x)=-\dfrac{p}{x^{p+1}}$. Your inequality becomes $$f'(x+1) \ge f(x+1)-f(x) = \dfrac{f(x+1)-f(x)}{(x+1)-x}$$
This follows easily from that $f$ is convex when $p>0$.
Edit: Since $f$ is convex, then for any $0<t<1$, $$f(x+t)=f((1-t)x+t(x+1))<(1-t)f(x)+tf(x+1) \implies \dfrac{f(x+1)-f(x+t)}{1-t}>f(x+1)-f(x)$$
Let $t \rightarrow 1^-$ and you get the result.
