the variance of the stochastic gradient computed by the i_th client is bounded by $$ \sigma^2 $$ - statistics

Question

The stochastic gradient $$ \nabla F_i (\omega, \Eta_i) $$ computed by the i_th client at model parameter $$ \omega $$ with minibatch $$Eta $$ has variance bounded by $$\sigma^2 $$.
E [ || \sum_{k=0}^{K-1} g_{i,k}^r ||^2 ] \leq K sum_{k=0}^{K-1} E [|| \nabla F_i (\omega_{r,k}^i) ||^2] + (K L^2 \rho ^2 \sigma_l ^2)/ N $$
The proof of the statement above was already done in the paper. I just couldn't understand the point of it. I would appreciate any hint on how the proof was completed. The proof can be found below:
$$ E [ || \sum_{k=0}^{K-1} g_{i,k}^r ||^2 ] = E [ || \sum_{k=0}^{K-1} g_{i,k}^r ||^2 ] + E[ || sum_{k=0}^{K-1} (g_{i,k}^r - \nabla F(w_{i,k}^r)) ||^2 ]
\leq K \sum_{k=0}^{K-1} E [|| g_{i,k}^r ||^2] + L^2 \sum_{k=0}^{K-1} E[ || 1/N \sum (\omega_{i,k}^r + \delta_{i,k}^r (\tilde(\omega_{i,k}^r); \Eta_{i,k}^r) - \omega_{i,k}^r - \delta_{i,k}^r (\tilde(\omega_{i,k}^r) ) ||^2 ]
\leq K \sum_{k=0}^{K-1} E [|| g_{i,k}^r ||^2] (KL^2 \rho ^2 \sigma_l ^2)/ N $$
I don't understand how the first step of the proof was made