I found this exercise on the internet ( I translated it from French so sorry if it's scuffed. ) I have no idea how to start it, any hint would be appreciated.
Let $(X_n)$ be a Markov chain with $Q$ being its transition matrix.
Let $E$ be the set of values the process take and let $x,y\in E$.
Prove that $\sum_{n=0}^{\infty}Q^n(y,x)\le \sum_{n=0}^{\infty}Q^n(x,x)$
EDIT based on @Michh's answer :
So I kinda get the general idea now but I didn't get these two transitions in his proof :
1/which conditional expectation property did we use get from line 1 to line 2 here? : $$\begin{align} \mathbb{E}_y[V_x] &= \mathbb{E}_y[V_x \mathbf{1}_{\{T_x<\infty\}}]\\ &= \mathbb{E}_y[ \mathbf{1}_{\{T_x<\infty\}}\mathbb{E}_y[V_x|\mathcal{F}_{T_x}]]. \end{align}$$
2/Also I dont get how we switched from conditional $y$ to conditional $x$ here $\mathbb{E}_y[V_x|\mathcal{F}_{T_x}]=\mathbb{E}_x[V_x]$. here I thought Markov's strong property only gives us independence so I get that we can get rid of $\mathcal{F}_t$
Denote by $V_x = \sum_{n=0}^\infty \mathbf{1}_{\{X_n = x\}}$ the number of visits of state $x$. Then what you want to prove is $$ \mathbb{E}_y [V_x] \leq \mathbb{E}_x[V_x].$$
To do this the idea is to apply the strong Markov property at the hitting time of $x$: $$T_x = \inf\{n\geq 1\colon \, X_n = x\}.$$
First notice that on $\{T_x = \infty\}$, we have $V_x = 0$ $\mathbb{P}_y$-a.s. Then $$\begin{align} \mathbb{E}_y[V_x] &= \mathbb{E}_y[V_x \mathbf{1}_{\{T_x<\infty\}}]\\ &= \mathbb{E}_y[ \mathbf{1}_{\{T_x<\infty\}}\mathbb{E}_y[V_x|\mathcal{F}_{T_x}]]. \end{align}$$ By the strong Markov property, $\mathbb{E}_y[V_x|\mathcal{F}_{T_x}] = \mathbb{E}_x[V_x].$ This gives $$\mathbb{E}_y[V_x] = \mathbb{P}_y(T_x<\infty)\mathbb{E}_x[V_x]\leq \mathbb{E}_x[V_x].$$