Need help understanding the last step of the proof of this lemma involving probabilities

Lemma :

Suppose that the events $A_{1}, ldots, A_{n}$ satisfy $operatorname{Var}left(sum_{i=1}^{n} I_{A_{i}}right) leq c sum_{i=1}^{n} Pleft{A_{i}right} .$ Then :
$$ left(1-Pleft(bigcup_{k=1}^{n} A_{k}right)right)^{2} sum_{i=1}^{n} Pleft{A_{i}right} leq c Pleft(bigcup_{k=1}^{n} A_{k}right)$$

Proof :

First off, set :
$$A =bigcup_{k=1}^{n} A_{k},,, beta=1-Pleft(Aright)$$

Without loss of generality, we can assume that $beta > 0$, if $beta = 0$ then the inequality holds trivially. Note that $A_i subset A$ for each $i = 1, …, n$, so $I_{A_i} I_A = I_A$. Also, $1 – beta = P(A)$.

begin{multline*}
sum_{j=1}^{n} Pleft(A_{j}right) =(1-P(A)) sum_{j=1}^nP(A_j) + P(A)sum_{j=1}^n P(A_j) \
= sum_{j=1}^n left[P(A_j) – P(A_j)P(A)right] + P(A)sum_{j=1}^n P(A_j)\
= sum_{j=1}^n E[I_{A_j}I_A – P(A_j)I_A)] + P(A)sum_{j=1}^n P(A_j)\
= Eleft[sum_{j=1}^n (I_{A_j} – P(A_j))I_A)right] + sum_{j=1}^n P(A_j)(1-beta) \
=Eleft{sum_{j=1}^{n}left(I_{A_{j}}-Pleft(A_{j}right)right) I(bigcup_{i=1}^{n} A_{i})right}+sum_{j=1}^{n} Pleft(A_{j}right)(1-beta) \
leqleft(operatorname{Var}left(sum_{j=1}^{n} I_{A_{j}}right) Pleft(bigcup_{j=1}^{n} A_{j}right)right)^{1 / 2}+sum_{j=1}^{n} Pleft(A_{j}right)(1-beta) \
leqleft(cleft(sum_{j=1}^{n} Pleft(A_{j}right)right) Pleft(bigcup_{j=1}^{n} A_{j}right)right)^{1 / 2}+sum_{j=1}^{n} Pleft(A_{j}right)(1-beta) \
color{red}{leq frac{c}{2 beta} Pleft(bigcup_{j=1}^{n} A_{j}right)+left(frac{beta}{2}+1-betaright) sum_{j=1}^{n} Pleft(A_{j}right)}\
end{multline*}

I understand they used the property $EI_A= P(A)$, the main assumption and Cauchy-Schwarz inequality to reach the before the last one step, but I can’t seem to figure how they went from that one to the last one (in red), any help will be greatly appreciated, thanks !