Suppose you have a statement $P(x,n)$ that's true for all reals $x$ and integers $n$, would this imply that $P(n,n)$ is true for all integers $n$?
For example, the Bernoulli's inequality usually states that $(1 + x)^n\geq 1 + nx$ for all real numbers $x\geq-1$ and all integers $n\geq 0$. To prove it, one may fix $x$, and then proceed it by induction on $n$. Now, the question is, is it allowed to let $x=n$, i.e. does it imply that $(1 + n)^n\geq 1 + n^2$ for all integers $n\geq 0$? If yes, why? (It may seem obvious, but not to me.)
hint
As you know, an integer is a real.
so, if some predicate is true for all reals, it will be true, in particular, for all integers.
Thus $$\forall (x,n)\in \Bbb R\times \Bbb N \;\; P(x,n) \implies $$ $$\forall (m,n)\in \Bbb N^2 \;\; P(m,n)$$
what about the case $m=n$.