#### Answer

See below.

#### Work Step by Step

Step (1) applies the fact that
$(-x)^{2}=(-1\cdot x)^{2}=(-1)^{2}x^{2}=1\cdot x^{2}=x^{2}$
(the square of a number is nonnegative, whether the number is negative or not, and, the squares of a number and its opposite number are equal in magnitude.)
So, given that $|a+b|$ is always nonnegative and
$(a+b) $ could be either negative or nonnegative,
their squares will be equal in any case.
Step (2) applies the fact that the product $ab$ could be negative or nonnegative.
If it is negative, it is less than $|a|\cdot|b|$, a product of nonnegative numbers.
If it it is nonnegative, it equals $|a|\cdot|b|$, a product of nonnegative numbers.
In either case $ab\leq|a|\cdot|b|$, so adding $2ab$ to $a^{2}+b^{2}$, we get less than or equal to the case where we add $2|a|\cdot|b|$ to $a^{2}+b^{2}$
Step (3) applies the same principle as step 1,
$|a|^{2}=a^{2}$ and $|b^{2}|=b^{2}$.
Step(4) Taking the square root of both sides, with both sides being nonnegative, we apply the inequality $''$LHS$\leq$RHS$''$ , achieved in step (2), because
$x_{1}\leq x_{2}\Rightarrow\sqrt{x_{1}}\leq\sqrt{x_{2}}$, for positive numbers $x_{1}$ and $x_{2}.$
(The square root function preserves order.)