Why does $(2n+1)^{d}-(2n-1)^{d}\leq d(2n+1)^{d-1}$ hold? Why does the inequality $(2n+1)^{d}-(2n-1)^{d}\leq d(2n+1)^{d-1}$ hold for $d,\;n \in \mathbb N$? I am sure I could do the above via induction. Nonetheless, I would like to know whether there are other ways to go about this without using induction?
 A: Something's wrong, because plugging in $n=1$, $d=2$ I get
$$3^2 - 1^2 \le 2 \cdot 3^1
$$
which says $8 \le 6$. Oops.
On the other hand, with another factor of $2$ on the right hand side it seems to be true. Substituting $a=2n+1$ and $b=2n-1$ into the formula
$$a^d-b^d=(a-b)(a^{d-1} + a^{d-2}b + ... + a b^{d-2} + b^{d-1})
$$
I get
\begin{align*}
(2n+1)^d - (2n-1)^d &= 2 \cdot \bigl((2n+1)^{d-1} + (2n+1)^{d-2} (2n-1) + ... \\
 & \qquad\qquad+ (2n+1)(2n-1)^{d-2} + (2n-1)^{d-1}\bigr) \\
 &\le 2d (2n+1)^{d-1}
\end{align*}
A: Because, by the mean value theorem, $$f(x+2)-f(x)=2f’(x_0)$$ for some $x_0\in[x,x+2].$
When $f’$ is increasing, as $f(x)=x^d,$ then:
$$2f’(x)\leq f(x+2)-f(x)\leq 2f’(x+2)$$
Letting $f(x)=x^d$ and $x=2n-1$ then:
$$2d(2n-1)^{d-1}\leq (2n+1)^d-(2n-1)^d \leq 2d(2n+1)^{d-1}$$
If instead you choose $f(x)=(2x-1)^d,$ you don’t get the multiple of $2$ in the mean value theorem, but you get the multiple of $2$ in the chain rule calculation of $f’(x).$

For your equality to be true, a necessary condition is:
$$1+\frac2{2n-1}=\frac{2n+1}{2n-1} \geq \sqrt[d-1]2$$
or
$$ n \leq \frac12 +\frac1{\sqrt[d-1]2-1}\tag{1}$$
Given any $d,$ (1) is true for only finitely many $n$. And even then, your inequality isn’t necessarily true, it’s only possibly true.
There are cases where your inequality is true. For $d\geq 3$ and $n=1,$ for example, or more generally, if $d\geq 2n+1.$
