Q:

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 $\operatorname{Var}\left(\sum_{i=1}^{n} I_{A_{i}}\right) \leq c \sum_{i=1}^{n} P\left\{A_{i}\right\} .$ Then :
$$ \left(1-P\left(\bigcup_{k=1}^{n} A_{k}\right)\right)^{2} \sum_{i=1}^{n} P\left\{A_{i}\right\} \leq c P\left(\bigcup_{k=1}^{n} A_{k}\right)$$
Proof :
First off, set :
$$A =\bigcup_{k=1}^{n} A_{k},\,\, \beta=1-P\left(A\right)$$
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} P\left(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)\\
= E\left[\sum_{j=1}^n (I_{A_j} - P(A_j))I_A)\right] + \sum_{j=1}^n P(A_j)(1-\beta)  \\
=E\left\{\sum_{j=1}^{n}\left(I_{A_{j}}-P\left(A_{j}\right)\right) I(\bigcup_{i=1}^{n} A_{i})\right\}+\sum_{j=1}^{n} P\left(A_{j}\right)(1-\beta) \\
 \leq\left(\operatorname{Var}\left(\sum_{j=1}^{n} I_{A_{j}}\right) P\left(\bigcup_{j=1}^{n} A_{j}\right)\right)^{1 / 2}+\sum_{j=1}^{n} P\left(A_{j}\right)(1-\beta) \\
 \leq\left(c\left(\sum_{j=1}^{n} P\left(A_{j}\right)\right) P\left(\bigcup_{j=1}^{n} A_{j}\right)\right)^{1 / 2}+\sum_{j=1}^{n} P\left(A_{j}\right)(1-\beta) \\
 \color{red}{\leq \frac{c}{2 \beta} P\left(\bigcup_{j=1}^{n}  A_{j}\right)+\left(\frac{\beta}{2}+1-\beta\right) \sum_{j=1}^{n} P\left(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 !

A:

This is a special case of Young's inequality, which states that for any constant $a > 0$,
$$ xy \leq \frac{x^2}{2a} + a \frac{y^2}{2}, $$
so if you use it on your product inside the paranthesis with $x = \sqrt{c \sum_j P(A_j)} , y = \sqrt{P(A)}$, and $a = \beta$, you can move to the last line.

