Help me understand quantifiers in logic Why the following statement is correct $\color{blue}{[\forall  x P(x) \lor \forall x Q(x) ] \Rightarrow \forall x[P(x) \lor  Q(x) ]}$ ? An explaination is given here, but I could not understand it completely.

Explanation says : let $P(x)$ mean x is rich and $Q(x)$ mean x is happy, then LHS says everyone is rich or everyone is happy, and RHS says everyone is rich or happy. Thus if LHS is true, then RHS is certainly true. However on RHS it can happen that two people are rich but the third is not rich but happy. In that case LHS is not true while RHS is true. Thus RHS does not necessarily imply LHS.

And What can be said about the statement $\color{blue}{[\forall  x P(x) \lor \forall x Q(x) ] \Leftrightarrow \forall x[P(x) \lor  Q(x) ]}$ and 
statement $\color{blue}{\forall x[P(x) \lor  Q(x) ] \Rightarrow [\forall  x P(x) \lor \forall x Q(x) ] }$ ?
 A: 
Why the following statement is correct $\color{blue}{[\forall  x P(x) \lor \forall x Q(x) ] \Rightarrow \forall x[P(x) \lor  Q(x) ]}$ ? An explaination is given here, but I could not understand it completely.

It may help to make the domain of quantification explicit. Let it be the set $U$. Then the problem can be restated as having to prove:
$$\forall  x: [x\in U\to  P(x)] \lor \forall x:[x\in U\to Q(x) ] \to \forall x:[x\in U \to P(x) \lor  Q(x)]$$
We can then begin our proof by assuming:
$\forall  x: [x\in U\to  P(x)] \lor \forall x:[x\in U\to Q(x)]$ 
Then suppose that we have $a\in U$. 
From our initial assumption we have two cases to consider:
Case 1: Assume $\forall  x: [x\in U\to  P(x)]$. 
For $x=a$, we have $P(a)$. From this, can infer that $P(a) \lor Q(a)$.
Case 2: Assume $\forall  x: [x\in U\to  Q(x)]$. 
For $x=a$, we have $Q(a)$. From this, can infer that $P(a) \lor Q(a)$.
In both cases, we have $P(a) \lor Q(a)$. Therefore, we can conclude and generalize, as required, that $$\forall x:[x\in U \to P(x) \lor  Q(x)]$$
A: $[\forall  x P(x) \lor \forall x Q(x) ]$ means EVERYBODY is happy, or EVERYBODY is is rich or EVERYBODY is both happy and rich.
$\forall x[P(x) \lor  Q(x) ]$ means that everybody is AT LEAST happy or rich, but there can be some only rich people and some only happy people and not everybody has tom be rich and not everybody has to be happy. This includes also the LHS.
A: $\color{blue}{[\,\forall  x~P(x) \lor \forall x~Q(x)\;] \implies \forall x~[P(x) \lor  Q(x) ]}$
"If either everything is P, or everything is Q, then everything is either P or Q."
Is it not clear that is at least one of the universals on the LHS holds then that on the RHS will hold?  Whatever P,Q mean.
If everything is P then everything is P or Q, elseif everything is Q, then everything is P or Q.  So, in either case, everything is P or Q.

And What can be said about the statement $\color{blue}{[\forall  x P(x) \lor \forall x Q(x) ] \Leftrightarrow \forall x[P(x) \lor  Q(x) ]}$ and
statement $\color{blue}{\forall x[P(x) \lor  Q(x) ] \Rightarrow [\forall  x P(x) \lor \forall x Q(x) ] }$ ?

Well, $\color{blue}{[\,\forall  x~P(x) \lor \forall x~Q(x)\;] \impliedby\forall x~[P(x) \lor  Q(x) ]}$ does not necessarily hold.
"If everything is either P or Q, then either everything is P, or everything is Q."
It is not necessary that if the RHS holds then one of the universal on the LHS will.
A: Consider a set with 2 elements:
x satisfies P(x) but not Q(x)
x' satisfies Q(x) but not P(x)
Then RHS is true: x satisfies P(x) so it satisfies $P(x) \vee Q(x)$, and x' satisfies Q(x) so it satisfies $P(x) \vee Q(x)$
Yet LHS is false: $\forall x P(x)$ is false because x' does not satisfies it, and $\forall x Q(x)$ is false because x does not satisfies it
