Let $N\in\mathbb{N}$ with $N\ge 2$ and $s, p\in\mathbb{R}$ with $s, p>1, p^*=Np/(N-p)$. Let $(x_n)_n$ be a sequence such that $x_n\to x_0$. Consider the quantity $$\frac{1}{(s+1)^{p-1}}(x_n^{p(s+1)}-x_0^{p(s+1)})- (x_n^{p^*(s+1)}-x_0^{p^*(s+1)}).$$ During the math class we said that the above quantity is equal to $$p(s+1)\left(\frac{\sigma_n^{p(s+1)-1}}{(s+1)^{p-1}}-\frac{N\tau_n^{p^*(s+1)-1}}{N-p} \right)(x_n-x_0)$$ for some $\sigma_n, t_n$ between $x_0$ and $x_n$ by the intermediate value theorem.
Could someone please explain me why it is true?
Thank you in advance.
Are you sure you mean the intermediate value theorem and not the mean value theorem?
If you define $F_1(x)=x^{p(s+1)}$ and $F_2(x)=x^{p^*(s+1)}$, then these functions are differentiable assuming $N\geq p$ in any interval. In particular, for any $n$ there exists $\sigma_n\in (x_0 ,x_n)$ and $\tau_n\in (x_0,x_n)$ such that
$$ F_1'(\sigma_n)\cdot(x_n-x_0)= \sigma_n^{p(s+1)-1} \cdot p(s+1)$$ and $$ F_2'(\tau_n)\cdot(x_n-x_0)= \tau_n^{p^*(s+1)-1} \cdot p^*(s+1)= \tau_n^{p^*(s+1)-1}\cdot p(s+1) \cdot \frac{N}{N-p}.$$