Prove that $\left(p-x\right)\left(x+1\right)^p+x^{p+1}\geq0$ I am trying to prove that, for all non-negative integers $x$ and all non-negative real numbers $p$,
$$
\left(p-x\right)\left(x+1\right)^p+x^{p+1}\geq0.
$$
I've been at this for a while and I'm stuck. I've tried finding positive functions smaller than this to compare it to, but no luck so far. If $p$ was an integer I might be able to do something with binomial coefficients, but I'm trying to solve for the general case.
 A: I got it!
The proof is rather long and there is probably an easier way, but I believe this is all correct.
Consider the function $f(x)=(1+x)^{-p}+px$. I start by showing that this function is never less than $1$ when $x$ and $p$ are non-negative. First, note:
$$
f(0)=1.
$$
Now, take the derivative of $f$:
$$
\begin{align}
f'(x) &= -p(1+x)^{-p-1}+p\\
&= p\left(1-\left(\frac1{1+x}\right)^{p+1}\right)\\
&\geq p(1-1)\\
f'(x)&\geq0.
\end{align}
$$
Since $f(0)=1$ and $f$ is never decreasing for positive $x$, $f(x)$ must be greater than or equal to $1$.
From here, we just have to do a bunch of rearranging.
$$
\begin{align}
(1+x)^{-p}+px &\geq 1\\
(1+x)^{-p} &\geq 1 - px\\
1 &\geq (1+x)^p(1-px)\\
\end{align}
$$
Now, since every positive $x$ has a corresponding positive $\frac1x$, we substitute $\frac1x$ for all instances of $x$
$$
\begin{align}
1 &\geq \left(1+\frac1x\right)^p\left(1-\frac px\right)\\
x\cdot x^p &\geq \left(x+1\right)^p\left(x-p\right)\\
x^{p+1} - \left(x+1\right)^p\left(x-p\right) &\geq 0\\
\left(p-x\right)\left(x+1\right)^p + x^{p+1} &\geq 0\\
\end{align}
$$
A: For $x,p\ge0$,
$$
\begin{align}
\left(1-\frac1{x+1}\right)^{p+1}&\ge1-\frac{p+1}{x+1}\tag1\\
\color{#C00}{((x+1)-1)^{p+1}}&\ge\color{#090}{(x+1)^{p+1}-(p+1)(x+1)^p}\tag2\\[6pt]
\color{#090}{((p+1)-(x+1))(x+1)^p}+\color{#C00}{x^{p+1}}&\ge0\tag3\\[6pt]
(p-x)(x+1)^p+x^{p+1}&\ge0\tag4\\
\end{align}
$$
Explanation:
$(1)$: Bernoulli's Inequality
$(2):$ multiply by $(x+1)^{p+1}$
$(3)$: $(x+1)-1=x$ on the left side
$\phantom{\text{(3):}}$ move $(x+1)^{p+1}-(p+1)(x+1)^p$ from the right side
$(4)$: $(p+1)-(x+1)=p-x$
A: Equivalently, we need to prove
\begin{align*}
&x \cdot x^p \ge (x - p) \cdot (x - p) \cdot (x + 1)^p\\
&\Longleftrightarrow \left(1 + \frac1x\right)^p \le \frac{x}{x - p}.
\end{align*}
But
\begin{align*}
\left(1 + \frac1x\right)^p &= 1 + \frac1x\binom{p}{x} + \frac1{x^2}\binom{p}{2} + \cdots\\
& \le 1 + \frac{p}{x} + \frac{p^2}{x^2} + \cdots\\
&= \frac{1}{1 - \frac px} = \frac{x}{x - p}.
\end{align*}
