Let $H=(V,E)$ be a $k$-uniform hypergraph with $n$ vertices, that is, $V:=V(H)$ is a $n$-element finite set of vertices and $E:=E(H)\subset\binom{V}{k}$ is a family of $k$-element subsets of $V$. 


Given $d$ vertices $v\_1,v\_2,\ldots,v\_d\in V(H)$, $1\leq d\leq k-1$, we denote by $deg\_H(v\_1,v\_2,\ldots,v\_d)$ the degree of the $d$-tuple $\{v\_1,v\_2,\ldots,v\_d\}$ in $H$, that is, the number of edges of $H$ which contain $v\_1,v\_2,\ldots,v\_d$. 


Further, let
$$\delta\_d(H):=\delta\_d=\min\{deg\_H(v\_1,v\_2,\ldots,v\_d):\{v\_1,v\_2,\ldots,v\_d\}\subset V(H)\}.$$


How to prove that
$$\delta\_{d-1}(H)\geq\frac{n-d+1}{k-d+1}\times\delta\_d(H).$$



Inequality about the minimum vertex degree in $k$-uniform hypergraphs
Double counting (as the form of the answer more or less says it has to be). Fix a $(d-1)$-set $X$; sum over vertices $v\not\in X$ the number of edges containing $X$ and $v$. You obviously count every edge containing $X$, and each edge is counted $k-d+1$ times. Since the sum is over $n-d+1$ vertices not in $X$, plugging in the minimum $d$-degree as a lower bound gives the desired inequality.


