Prove that $\left(p-x\right)\left(x+1\right)^p+x^{p+1}\geq0$

Question

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.

Answer 1

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 $x$ $$ \begin{align} 1 &\geq \left(1+\frac1x\right)^p\left(1-\frac px\right)\\ x\cdot x^p &\geq \left(x+1\right)^p\left(x-p\right)\\ x^{p+1} - \left(x+1\right)^p\left(x-p\right) &\geq 0\\ \left(p-x\right)\left(x+1\right)^p + x^{p+1} &\geq 0\\ \end{align} $$

Answer 2

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
$(4)$: $(p+1)-(x+1)=p-x$

Answer 3

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*}