A very elementary inequality with polynomials Let $s,t \in \mathbb{R}$, and let $j \in \{1, 2, \dots,N\}$. Suppose that $s \ge t$ and that $s^j - t^j \ge s - t$. Is it true that $s^N - t^N \ge s^j - t^j$?
Thank you for your time and help!
EDIT: taking into account the good, but negative, answres I've received, let me explain the original problem!
Let $f(u) = \sum_{j=0}^{N} b_ju^j$. I am trying to estimate $|f(u) - f(v)| = |\sum_{j=1}^N b_j(u^j - v^j)|$ in such a way as to obtain $|u - v| \cdot(\text{something})$. 
I was hesitant to apply Lagrange's theorem because $u,v \in H^1_0(\Omega)$, where $\Omega$ is a bounded domain in $\mathbb{R}^d$, $d = 1,2$ such that $||u||_{H^1_0}\ ,\ ||v||_{H^1_0} \le R$. 
 A: It's not true even when $s$ and $t$ are positive.  Take $j=2$, then for any $s > t \ge 1/2$ we have $s^2 - t^2 = (s-t)(s+t) > s-t$.
Now further assume $s < 1$ so that $s^N \to 0$ (and $t^N \to 0$).  For all large $N$ (how large depends on $s$ and $t$) we'll have $s^N - t^N < s^2-t^2$.
A: As stated the conclusion does not follow.
Let $s=1$ and $t=-1$.  Then for example, with $j=3$ we have $s^j - t^j = s - t = 2$, but taking $N=4$ produces $s^N - t^N = 0 $.
A: if $s \ge 1 \ge t\ge 0$, then for any $j\in\{1,\cdots,N\}$
$$
s^N-t^N = (s-t)(s^{N-1}+s^{N-2}t+\cdots+st^{N-2}+t^{N-1})
\\ = (s-t)[s^{N-j}(s^{j-1}+s^{j-2}t+\cdots+st^{j-2}+t^{j-1})+s^{N-j-1}t^{j}+\cdots+st^{N-2}+t^{N-1}]
\\ = s^{N-j}(s^j-t^j)+(s-t)(s^{N-j-1}t^{j}+\cdots+st^{N-2}+t^{N-1})
\\ \ge s^{N-j}(s^j-t^j)
\\ \ge s^j-t^j
$$
since the second term is nonnegative. in particular, the condition
$$
s^j-t^j \ge s-t
$$
always holds (substitute $j$ for $N$ and $1$ for $j$). if $0<s<1$ is allowed, then
$$
s^{-N}(s^N-t^N) \ge s^{-j}(s^j-t^j)
\\ s^{-j}(s^j-t^j) \ge s^{-1}(s-t)
$$
A: I finally solved also the problem in the edit:
$f(u) - f(v) = (u - v) \sum_{j=1}^{N}b_j \sum_{i=0}^{j-1}u^{j-1-i}v^i$ and everything follows...
