My question is stated in the title: If $Y_n \xrightarrow{p}Y$, can we use triangle inequality $|Y_n - Y_{n+1}| \le |Y_n - Y| + |Y_{n+1} - Y|$ to prove that $Y_{n+1} - Y_n$ goes to 0 in probability?
When I saw this on a material, it just simply declare that is true. But when I look into the definition of convergence probability, I found it is not that simple.
Definition for convergence in probability: $$ \forall \epsilon>0, \lim _{n \rightarrow \infty} P\left(\left|Y_{n}-Y\right|<\epsilon\right)=1 $$ So to show the convergence of $Y_{n+1} - Y_n$, we need to show $P(|Y_n - Y_{n+1}| < \epsilon) \ge P(|Y_n - Y| < \epsilon) + P(|Y_{n+1} - Y| < \epsilon)$.
I tried the following procedure: since $|Y_n - Y_{n+1}| \le |Y_n - Y| + |Y_{n+1} - Y|$, then $$ P(|Y_n - Y_{n+1}| < \epsilon) \ge P(|Y_n - Y| + |Y_{n+1} - Y| < \epsilon) \ge P(|Y_n - Y|< \epsilon/2 \cup |Y_{n+1} - Y| < \epsilon/2)\\ \le P(|Y_n - Y|< \epsilon/2 ) + P(|Y_{n+1} - Y| < \epsilon/2)) $$ But last inequality is less or equal, so we cannot prove it by that. And I don't know how to continue. Please advice.
Hint: $|Y_n-Y_{n+1}| >\epsilon$ implies that either $|Y-Y_{n+1}| >\epsilon /2$ or $|Y-Y_{n}| >\epsilon /2$ (from the inequality you wrote). Can you finish?