I think considering this is the university help section, it is fair to say this proof is not complicated. It just relies on the properties of the real numbers.
$\displaystyle x\leq |x|$ follows from the fact that the reals are ordered.
$\displaystyle a \in \mathbb{R}, b \in \mathbb{R} \implies ab \in \mathbb{R}$ depends on how you look at R, easiest reasoning is because R is a field.
The rest follows from simple algebra.
The statements of the proof are :
1) $\displaystyle ab\leq |ab|$
2) $\displaystyle 2ab\leq 2|a||b|$
3) $\displaystyle (a+b)^2\leq (|a|+|b|)^2$
4) $\displaystyle |a+b|\leq |a|+|b|$
And the problem ask us :
a) To mention appropriate theorems or definitions that are responsible for those statements.
b) How are they involved .
For the (a) part the following tabular form shows clearly the theorems or definitions involved in the proof.
1)$\displaystyle ab\leq |ab|$.................................................. ............by using the theorem : for all ,x:$\displaystyle x\leq |x|$
2)|ab| = |a||b|............................................ ..................................by using the theorem : for all , x, y : |xy| = |x||y|
3)$\displaystyle ab\leq |a||b|$.................................................. .........by substituting (2) into (1)
4)$\displaystyle 2ab\leq 2|a||b|$.................................................. .......by using the theorem : for all ,x,y,z : z>0 and $\displaystyle x\leq y\Longrightarrow xz\leq yz$
5)$\displaystyle a^2+b^2+2ab\leq 2|a||b|+a^2+b^2$.............................by using the theorem : for all ,x,y,z : $\displaystyle x\leq y\Longleftrightarrow x+z\leq y+z$
6)$\displaystyle a^2 = |a|^2$.................................................. ............by using the theorem :for all ,x : $\displaystyle x^2 = |x|^2$
7)$\displaystyle b^2 = |b|^2$.................................................. ............by using again the same theorem as in (6)
8)$\displaystyle a^2+b^2+2ab\leq 2|a||b|+|a|^2+|b|^2$.................................................. ..........by substituting (7) and (6) into (5)
9)$\displaystyle (a+b)^2 = a^2+b^2+2ab$............................................by using the identity (theorem) : for all , x,y :$\displaystyle (x+y)^2 = x^2+y^2+2xy$
10)$\displaystyle (|a|+|b|)^2 = |a|^2+|b|^2+2|a||b|$...........................by using the same theorem as in (9)
11)$\displaystyle (a+b)^2\leq (|a|+|b|)^2$............................................by substituting (10) and (9) into (8)
12)$\displaystyle \sqrt {(a+b)^2}\leq\sqrt {(|a|+|b|)^2}$........................by using the theorem : for all ,x,y : $\displaystyle x\geq 0$ and $\displaystyle y\geq 0\Longrightarrow (x\leq y\Longleftrightarrow\sqrt x\leq\sqrt y)$
13)$\displaystyle \sqrt{(a+b)^2} = |a+b|$..............................................by using the theorem : for all ,x : $\displaystyle \sqrt{ x^2} = |x|$
14)$\displaystyle \sqrt{(|a|+|b|)^2} = ||a|+|b||$.............................................by using the same theorem as in (13)
15)$\displaystyle |a+b|\leq ||a|+|b||$.................................................. ..........by substituting (14) and (13) into (12)
16) ||a|+|b|| = |a|+|b|........................................... .........................by using the definition of absolute value : for all , x : $\displaystyle x\geq 0\Longrightarrow |x| = x$
17) $\displaystyle |a+b|\leq |a|+|b|$.................................................. .by substituting (16) into (15).
Note: that in the above many other statements that are not mention in the proof are exposed . To include them all or part of them in the proof it is matter that depends on the writer's style .
NOW for part (b) .
Part (b) deals with the laws of logic ,the application of which, transforms the theorems involved in the proof into the statements of the proof.
That part i leave to Plato ,since he thinks that the whole problem is not complicated.
Of course if he wishes to do that