Why the $\textbf{PCP}(\log n, 1)$ construction of completeness $c$ and soundness $s$ for an $NP$ language $L$ implies $s/c$ inapproximability for $L$?
The authors give a reduction $x \mapsto f(x)$ from some NP-hard problem $X$ to MAX-CUT with the following properties:


If $x$ is a YES instance then $OPT(f(x)) \geq c$.
If $x$ is a NO instance then $OPT(f(x)) \leq s$.


Here $OPT(y)$ is the optimal value of the MAX-CUT instance $y$.

Suppose now that you had an algorithm for MAX-CUT with approximation ratio $\rho &gt; s/c$, which means that for every instance $y$, $ALG(y) \geq \rho OPT(y)$, where $ALG(y)$ is the value of the instance produced by the algorithm. We can use this algorithm to solve $X$, since $x$ is a YES instance of $X$ iff $OPT(f(x))&gt;s$ (I'll let you verify that).

This informally explains the statement in the paper. Any more formal explanation requires a definition of "NP-hard to approximate". One possibility is as follows. Let $Y$ be a maximization problem. We say that $Y$ is NP-hard to approximate within a factor of $\rho$ if for every problem $X \in \mathsf{NP}$ there exists a polytime computable reduction $f$ which on input $x$ produces a tuple $(y,v)$, where $y$ is an instance of $Y$ and $v$ is a number, with the following properties:


If $x$ is a YES instance of $X$ then $OPT(y) \geq v$.
If $x$ is a NO instance of $X$ then $OPT(y) &lt; \rho v$.


This generalizes the usual definition of NP-hardness for optimization problems, which corresponds to the choice $\rho = 1$.
Under this definition, if $Y$ is NP-hard to approximate within a factor of $\rho$ and there is a polytime $\rho$-approximation algorithm for $Y$, then $\mathsf{P} = \mathsf{NP}$. 
Finally, according to this definition, MAX-CUT is indeed NP-hard to approximate to within any factor strictly larger than $s/c$.