Trouble Understanding Proof Of Invariant Relationship In part of a proof I am reading this is stated:
$2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) + (a_n + c_n )^2 + (b_n + d_n )^2 ≥ 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ).$ (1)
From this invariant inequality relationship we conclude that, for $n ≥ 2$
$
a_n^2 + b_n^2 + c_n^2 + d_n^2 ≥ 2 ^{n - 1} (a_1^2 + b_1^2 + c_1^2 + d_1^2 ).$   (2)
I do not understand how (2) was derived from (1)?
Full proof:  Proof 

They made the substitution $(a_n + c_n )^2 + (b_n + d_n )^2 = -2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$
in the equality:
$$(a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2) \\= 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) - 2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$$
to write it as: 
$$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 = 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) + (a_n + c_n )^2 + (b_n + d_n )^2$$
Hence, $$\displaystyle a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) \text{ for each $n \ge 1$}$$
(since, $(a_n + c_n )^2 + (b_n + d_n )^2 \ge 0$)
inductively leads to: $$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2)$$
$$2(a_n^2 + b_n^2 + c_n^2 + d_n^2) \ge 4(a_{n-1}^2 + b_{n-1}^2 + c_{n-1}^2 + d_{n-1}^2)$$
$$\cdots$$
$$2^{n-1} (a_2^2 + b_2^2 + c_2^2 + d_2^2) \ge 2^{n} (a_1^2 + b_1^2 + c_1^2 + d_1^2)$$
Now add these $n$ inequalities !
