Can I use ": ," instead of ", implies" for this example? I've to write this statement in a formal manner:

if $x>1$ then $x^{2}>1$.

Writing the result of the exercise I face this problem, I wonder if these two statements are equivalent:
$$\forall~x \in \mathbb{R}, x >1 \implies x^{2}>1 \ $$
$$\forall~x \in \mathbb{R}: x >1, ~~ x^{2}>1 \ $$
I read the first as:

for all the x in $\mathbb{R}$ if x is greater than 1 then x squared is greater than 1

For the second I read it so:

for all the x in $\mathbb{R}$ such that x is greater then 1, x squared is greater then 1

Where is my fault? Is there a better alternative? 

Exercise 1.14-14 p.23 from Analisi matematica vol.1 3th ed. (2002) by Enrico Giusti 

These statemens aren't equivalent, at least the way I read them. I read the first as $$
  \forall x\,\left[\left((x \in \mathbb{R}) \land (x > 1)\right) \rightarrow x^2 > 1 \right] \text{,}
$$
which of course is true. The second, however, I read as $$
  \forall x \left[(x \in \mathbb{R}) \rightarrow \left((x > 1) \land (x^2 > 1)\right)\right] \text{,}
$$
which is wrong, since $-1 \in \mathbb{R}$ but $x < 1$.
I'd say the general rule for reading such first-order formulas with extended syntax is that $$
  \forall x\,\varphi_1,\varphi_2,\ldots,\varphi_n\,:\, \theta
$$ 
means $$
  \forall x \left[\left(\varphi_1 \land \cdots \land \varphi_n\right) \rightarrow \theta\right] \text{,}
$$
i.e. in plain english that "For all $x$ such that $\varphi_1$ and $\varphi_2$ and ... $\varphi_n$, we have $\theta$".
Your first formula doesn't include a colon (:), and therefore doesn't really fit into that schema. But since the last comma-separated term is an implication, i.e. since it has the form $$
  \forall x\,\varphi_1,\ldots,\varphi_n \rightarrow \theta 
$$
I'd say most people will read it as $$
    \forall x \left[\left(\varphi_1 \land \cdots \land \varphi_n\right) \rightarrow \theta\right] \text{.}
$$
But it really is a bit ambiguous, since another possible reading would be $$
    \forall x \left[\varphi_1 \land \cdots \land \varphi_{n-1} \land (\varphi_n \rightarrow \theta)\right] \text{.}
$$
