It’s the same process as for a simpler relation; the notation just gets a bit more cumbersome.
To show that $R$ is reflexive, you’d have to show that $(i,j)\,R\,(i,j)$ for every $(i,j)$. Now translate that into more basic terms, using the definition of $R$: you’d have to show that for each $(i,j)$, either
$$i or
$$i
Since these are the same statement, you'd have to prove that $i$(i,j)$. But you can do at most half of this: if $i\le j$, then it’s true that $i\le j\le j$, but it’s never true that $i. Thus, $R$ is not reflexive: it’s not true that $(i,j)\,R\,(i,j)$ for each $(i,j)$. In fact, it’s not true for any $(i,j)$, so $R$ is irreflexive.
I’m going to leave symmetry for you; it’s pretty easy if you think about the form of the definition of $R$, and transitivity is a bit messier.
To show that $R$ is transitive, you’d have to show that for any $(i,j),(k,\ell)$, and $(m,n)$, if $(i,j)\,R\,(k,\ell)$ and $(k,\ell)\,R\,(m,n)$, then $(i,j)\,R\,(m,n)$. Here again your automatic first step should be to translate this:
and the desired conclusion is
You could systematically investigate all of the combinations: $i $$i which is certainly possible. Thus, we can’t in general infer $(i,j)\,R\,(m,n)$ from $(i,j)\,R\,(k,\ell)$ and $(k,\ell)\,R\,(m,n)$: if $i=m (Actually, if you show first that $R$ is irreflexive and symmetric, you can take a shortcut to show that it can’t be transitive. Just find two different pairs $(i,j)$ and $(k,\ell)$ such that $(i,j)\,R\,(k,\ell)$. Then by symmetry $(k,\ell)\,R\,(i,j)$, so if $R$ were transitive, we could deduce that $(i,j)\,R\,(i,j)$. However, ... )