Question about step in Ahlfors' proof of Cauchy's inequality in complex analysis.

I’m reading Ahlfors’ Complex Analysis. In this book, he provides the following proof of Cauchy’s inequality. Using $|a -b|^2 = |a|^2 + |b|^2 – 2 Releft(aoverline{b}right)$ he establishes the following

$$0le sum_{k=1}^n bigrlvert a_k – lambda overline{b_k}bigrrvert^2 = sum_{k=1}^n |a_k|^2 + underbrace{|lambda|^2sum_{k=1}^n |b_k|^2}_{a)} – 2 underbrace{Releft(overline{lambda}sum_{k=1}^n a_kb_kright)}_{b)}$$

where $lambda$ is some arbitrary complex number. He then proceeds to take the particular value of $lambda$ to be

$$ lambda =frac{sum_{j=1}^n a_jb_j}{sum_{j=1}^n |b_j|^2} $$

and using this, he says that after simplifications you obtain the following:

$$ sum_{k=1}^n |a_k|^2 – frac{Bigrlvertsum_{k=1}^n a_kb_kBigrrvert^2}{sum_{k=1}^n |b_k|^2}ge0$$

which proves Cauchy’s inequality.

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I wanted to expand this and check this result for myself. I separated the problem into $2$ parts:

$textbf{a)}$ For $|lambda|^2sum_{k=1}^n |b_k|^2$ I got the follwing:
$$
|lambda|^2sum_{k=1}^n |b_k|^2 = frac{Bigrlvertsum_{j=1}^n a_jb_jBigrrvert^2}{Bigrlvertsum_{j=1}^n |b_j|^2Bigrrvert^2}sum_{k=1}^n |b_k|^2 = frac{Bigrlvertsum_{j=1}^n a_jb_jBigrrvert^2}{left(sum_{j=1}^n |b_j|^2right)^2}sum_{k=1}^n |b_k|^2 = frac{Bigrlvertsum_{k=1}^n a_kb_kBigrrvert^2}{sum_{k=1}^n |b_k|^2}
$$
Here I use the property $bigrlvert|x|^2+|y|^2bigrrvert = |x|^2 + |y|^2$. I believe this is justified because $|x|^2 + |y|^2in mathbb{R}$, that $|x|^2 + |y|^2ge 0 + 0 ge 0$, and that the modulus of a positive real number is the real number itself. I think the reasoning is correct, but I’m not completely sure.

$textbf{b)}$ For $Releft(overline{lambda}sum_{k=1}^n a_kb_kright)$ I got the following:
$$
Releft(overline{lambda}sum_{k=1}^n a_kb_kright) = Releft(frac{overline{sum_{j=1}^n a_jb_j}}{overline{sum_{j=1}^n |b_j|^2}}sum_{k=1}^n a_kb_kright) = Releft(frac{sum_{j=1}^n overline{a_jb_j}}{sum_{j=1}^n |b_j|^2}sum_{k=1}^n a_kb_kright) =frac{Releft( sum_{j=1}^nsum_{k=1}^noverline{a_jb_j}a_kb_kright)}{ sum_{j=1}^n |b_j|^2}
$$

Where I used the fact that $overline{x} = x$ for $x in mathbb{R}$, and also that $Releft(frac{x}{c} + i frac{y}{c}right) = frac{x}{c} = frac{Re(x + iy)}{c}$. And here is where I ran into trouble.

I know that for the specific case where $j=k$ I can make a simplification using the fact that $z overline{z} = |z|^2$, but this still leaves the other cases where $j neq k$, and I don’t know how I could find the real part of these terms.

I also tried using the fact that $Re(z) le |z|$ and that $|a +b| le |a| + |b|$. Using this I got that
$$
frac{Releft( sum_{j=1}^nsum_{k=1}^noverline{a_jb_j}a_kb_kright)}{ sum_{j=1}^n |b_j|^2} le frac{Bigrlvert sum_{j=1}^nsum_{k=1}^noverline{a_jb_j}a_kb_kBigrrvert}{ sum_{j=1}^n |b_j|^2} le frac{ sum_{j=1}^nsum_{k=1}^n|a_jb_j|cdot|a_kb_k|}{ sum_{j=1}^n |b_j|^2}le frac{ sum_{s=1}^n|a_sb_s|^2}{ sum_{j=1}^n |b_j|^2}
$$
where for the last inequality I just did $sum_{j=k} + sum_{jneq k} le sum_{j=k}$. But even with this, this doesn’t give me a result that simplifies to the desired conclusion.

I don’t if there’s a step I’m doing wrong or of there’s something I’m missing, but I can’t seem to get to the inequality I want to arrive at. Can anyone tell if I’m on the right track? Thank you!