I'm reading Ahlfors' Complex Analysis. In this book, he provides the following proof of Cauchy's inequality. Using $|a -b|^2 = |a|^2 + |b|^2 - 2 \Re\left(a\overline{b}\right)$ he establishes the following
$$0\le \sum_{k=1}^n \bigr\lvert a_k - \lambda \overline{b_k}\bigr\rvert^2 = \sum_{k=1}^n |a_k|^2 + \underbrace{|\lambda|^2\sum_{k=1}^n |b_k|^2}_{a)} - 2 \underbrace{\Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right)}_{b)}$$
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:
$$ \sum_{k=1}^n |a_k|^2 - \frac{\Bigr\lvert\sum_{k=1}^n a_kb_k\Bigr\rvert^2}{\sum_{k=1}^n |b_k|^2}\ge0$$
which proves Cauchy's inequality.
I wanted to expand this and check this result for myself. I separated the problem into $2$ parts:
$\textbf{a)}$ For $|\lambda|^2\sum_{k=1}^n |b_k|^2$ I got the follwing: $$ |\lambda|^2\sum_{k=1}^n |b_k|^2 = \frac{\Bigr\lvert\sum_{j=1}^n a_jb_j\Bigr\rvert^2}{\Bigr\lvert\sum_{j=1}^n |b_j|^2\Bigr\rvert^2}\sum_{k=1}^n |b_k|^2 = \frac{\Bigr\lvert\sum_{j=1}^n a_jb_j\Bigr\rvert^2}{\left(\sum_{j=1}^n |b_j|^2\right)^2}\sum_{k=1}^n |b_k|^2 = \frac{\Bigr\lvert\sum_{k=1}^n a_kb_k\Bigr\rvert^2}{\sum_{k=1}^n |b_k|^2} $$ Here I use the property $\bigr\lvert|x|^2+|y|^2\bigr\rvert = |x|^2 + |y|^2$. I believe this is justified because $|x|^2 + |y|^2\in \mathbb{R}$, that $|x|^2 + |y|^2\ge 0 + 0 \ge 0$, 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 $\Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right)$ I got the following: $$ \Re\left(\overline{\lambda}\sum_{k=1}^n a_kb_k\right) = \Re\left(\frac{\overline{\sum_{j=1}^n a_jb_j}}{\overline{\sum_{j=1}^n |b_j|^2}}\sum_{k=1}^n a_kb_k\right) = \Re\left(\frac{\sum_{j=1}^n \overline{a_jb_j}}{\sum_{j=1}^n |b_j|^2}\sum_{k=1}^n a_kb_k\right) =\frac{\Re\left( \sum_{j=1}^n\sum_{k=1}^n\overline{a_jb_j}a_kb_k\right)}{ \sum_{j=1}^n |b_j|^2} $$
Where I used the fact that $\overline{x} = x$ for $x \in \mathbb{R}$, and also that $\Re\left(\frac{x}{c} + i \frac{y}{c}\right) = \frac{x}{c} = \frac{\Re(x + iy)}{c}$. 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 $$ \frac{\Re\left( \sum_{j=1}^n\sum_{k=1}^n\overline{a_jb_j}a_kb_k\right)}{ \sum_{j=1}^n |b_j|^2} \le \frac{\Bigr\lvert \sum_{j=1}^n\sum_{k=1}^n\overline{a_jb_j}a_kb_k\Bigr\rvert}{ \sum_{j=1}^n |b_j|^2} \le \frac{ \sum_{j=1}^n\sum_{k=1}^n|a_jb_j|\cdot|a_kb_k|}{ \sum_{j=1}^n |b_j|^2}\le \frac{ \sum_{s=1}^n|a_sb_s|^2}{ \sum_{j=1}^n |b_j|^2} $$ where for the last inequality I just did $\sum_{j=k} + \sum_{j\neq k} \le \sum_{j=k}$. 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!
Hint: With the choice of $\lambda=\frac {\sum_{j=1}^n a_jb_j}{\sum_{j=1}^n|b_j|^2},$ $$\bar{\lambda}\sum_{k=1}^na_kb_k$$ is already real and equals $$\frac{|\sum_{j=1}^n a_j b_j|^2}{\sum _{j=1}^n|b_j|^2}.$$