Yes, indeed you need that $\gcd(x,N)=1$, because otherwise the statement is not true. Look e.g. at $2^r\equiv 1\pmod 4$, which is only true for $r=0$ but not for any positive integer. So I will assume $\gcd(x,N)=1$ in the following.
Look at the list
$$x^0,x^1,x^2,...,x^{N-1},x^N$$
modulo $N$. This can give you at most $N$ different numbers, namely $0,...,N-1$. But the list has length $N+1$, so some number must ocure at least twice (pidgeonhole principle).
So, let's say $x^n\equiv x^m\pmod N$ with $n>m$. Maybe you already know that for $\gcd(x,N)=1$ there is a unique multiplicative inverse $x^{-1}$ so that $x\cdot x^{-1}\equiv 1\pmod N$. We multiply $x^{-m}:=(x^{-1})^m$ to both sides:
\begin{align}
x^n&\equiv x^m\pmod N &|\cdot x^{-m} \\
x^n\cdot x^{-m}&\equiv x^m\cdot x^{-m}\pmod N \\
x^{n-m}&\equiv 1\pmod N \\
\end{align}
Since $n,m\le N$ and $n>m$ we have that $r:=n-m\le N$ (actually $r<N$).
Let $x\geq y$.
Thus, we need to probe that
$$\left(\sqrt[n]x-\sqrt[n]y\right)^n\leq x-y$$ or
$$\left(\sqrt[n]x-\sqrt[n]y+\sqrt[n]y\right)^n\geq\left(\sqrt[n]x-\sqrt[n]y\right)^n+\left(\sqrt[n]y\right)^n,$$ which is obvious for natural $n$.
Best Answer
Assume that $a$ and $b$ are positive integers satisfying $ab|(a+b)(a+b+1)$. Set $gcd(a,b)=x$, and set $a_x = a/x$, and $b_x=b/x$. Note that $xa_x = a$ and $xb_x=b$. Note also that $(a_x,b_x)=1.$ Since $$ab|(a+b)(a+b+1)$$ one has $$x^2a_xb_x|(xa_x + xb_x)(xa_x + xb_x +1)$$ which implies that $$xa_xb_x|(a_x + b_x)(xa_x + xb_x +1).$$ Now note that $gcd(x,xa_x + xb_x +1)=1$ so the previous relation forces $x|a_x+b_x$. We have then $$x \leq a_x + b_x = \frac{a}{x} + \frac{b}{x}.$$ One has then from clearing $x$ in the denominator $$x^2 \leq a+b$$ which implies the desired inequality.
Note that from a similar argument you can actually get a lower bound on $x$ and obtain that $$x \geq \sqrt{\frac{ab}{a+b+1}}.$$ So the actual possible range for the gcd is pretty tiny.
I'm highly curious where this problem came from. It isn't one I've seen before.