Let $X_1,X_2,\dots,X_n$ be arbitrary discrete random variables. Prove that $$H(X_1^n)\leq\frac{1}{n-1}\sum_{i=1}^nH(X_1^{i-1},X_{i+1}^n),$$ where for $j\geq i$, $X_i^j$ denotes the block of random variables $(X_i,\dots,X_j)$ and for $j<i$ $X_i^j$ can just be interpreted as an empty constant zero.
My idea was to show this using induction. The base case $n=2$ is obvious from the chain rule: $$H(X_1,X_2)=H(X_1)+H(X_2|X_1)\leq H(X_1)+H(X_2).$$ Then assuming true for $n=k$, the idea was to use the chain rule and hypothesis again: $$H(X_1^{k+1})=H(X_1^k)+H(X_{k+1}|X_1^k)\leq \frac{1}{k-1}\sum_{i=1}^kH(X_1^{i-1},X_{i+1}^k)+H(X_{k+1}|X_1^k).$$ But I'm unsure how to proceed from here because of the fraction making it difficult to merge the terms in a meaningful way. Any advice would be greatly appreciated!
This fact is often referred to as Han's Inequality. It's hard to use induction to prove this since like you've noticed it gets a bit messy from the $1/(k-1)$. Instead you can use the inequality, valid for all $i=1,...,n$
\begin{align} H(X_1^n) &= H(X_1^{i-1}, X_{i+1}^n) + H(X_i|X_1^{i-1}, X_{i+1}^n) \\ &\leq H(X_1^{i-1}, X_{i+1}^n) + H(X_i|X_1^{i-1}) \end{align} where the first line is the chain rule, and the second line is that conditioning reduces entropy. To finish up, see what happens when you sum both sides over $i=1,...,n$. As a hint, think about the value of this sum $$\sum_{i=1}^n H(X_i|X_1^{i-1}).$$