I don't know how to proof the following question:
$$py^{p-1}(x-y)<x^p – y^p<px^{p-1}(x-y),(0<y<x,p>1)$$
Thank you very much.
[Math] Prove that $py^{p-1}(x-y)1)$
calculusinequality
calculusinequality
I don't know how to proof the following question:
$$py^{p-1}(x-y)<x^p – y^p<px^{p-1}(x-y),(0<y<x,p>1)$$
Thank you very much.
Best Answer
As JPi says:
Let $f(x) = x^p$ with $p > 1$. By the mean value theorem we have $\frac{f(x)-f(y)}{x-y} = f'(c) $ where $y < c < x$.
We also have $f'(x) =px^{p-1}> 0$ and $f''(x) =p(p-1)x^{p-2}> 0$.
Since $f''(x) > 0$, $f'(y) < f'(c)=\frac{f(x)-f(y)}{x-y} < f'(x)$, which is what you want.
Note that if $0 < p < 1$, then $f''(x) < 0$, so the inequalities are reversed: $f'(y) > f'(c)=\frac{f(x)-f(y)}{x-y} > f'(x)$.