Proof of a real version of a discrete inequality Good evening, I was wondering how to prove easily this inequality: \begin{equation} py^{p-1}(x-y)\leq x^p-y^p\leq px^{p-1}(x-y) \end{equation} for all $x,y\geq0$ and $p\in[1,+\infty)$. I tried to study both the functions $\frac{x^p-y^p}{y^{p-1}(x-y)}$ and $\frac{x^p-y^p}{x^{p-1}(x-y)}$, but I'm not getting anywhere.
mean value theorem question I was trying to solve the following:  Given: $0 &lt; a &lt; b$ and $n&gt;1$, prove: $$na^{n-1}(b-a) &lt; b^n-a^n &lt; nb^{n-1}(b-a)$$  I managed to get this far using the mean value theorem: $$a^n(b-a)&lt;b^n -a^n&lt;b^n (b-a)$$ Any idea how to continue?
$x^a - 1$ divides $x^b - 1$ if and only if $a$ divides $b$ Let $x &gt; 1$ and $a$, $b$ be positive integers. I know that $a$ divides $b$ implies that $x^a - 1$ divides $x^b - 1$. If $b = qa$, then  $$x^b - 1 = (x^a)^q - 1^q = (x^a - 1)((x^a)^{q-1} + \ldots + x^a + 1).$$  I'm interested in the converse of the statement. If $x^a - 1$ divides $x^b - 1$, does this imply that $a$ divides $b$?