Prove that…

Prove that $\ \left|\displaystyle\sum_{k=1}^na_kb_k\right|^2\le\left(\displaystyle\sum_{k=1}^n|a_k|^2\right)\left(\displaystyle\sum_{k=1}^n|b_k|^2\right)\ $ for $a_k,b_k\in\mathbb C$.

Step 1 · Handle the trivial case

If $B:=\displaystyle\sum_{k=1}^n|b_k|^2=0$, then every $b_k=0$ and both sides of the inequality are $0$. Assume $B>0$.

Step 2 · Start from a nonnegative sum

For any $\lambda\in\mathbb C$:

$$ 0\le\sum_{k=1}^n|a_k-\lambda b_k|^2=\sum_{k=1}^n|a_k|^2-\lambda\sum_{k=1}^nb_k\bar a_k-\bar\lambda\sum_{k=1}^na_k\bar b_k+|\lambda|^2\sum_{k=1}^n|b_k|^2 $$

Step 3 · Introduce notation

Let $A=\displaystyle\sum_k|a_k|^2$, $\ B=\displaystyle\sum_k|b_k|^2$, $\ S=\displaystyle\sum_ka_k\bar b_k$. Then:

$$ 0\le A-\lambda\bar S-\bar\lambda S+|\lambda|^2B $$

Step 4 · Choose the optimal $\lambda$

Take $\lambda=\dfrac SB$ (so $\bar\lambda=\dfrac{\bar S}B$, $|\lambda|^2=\dfrac{|S|^2}{B^2}$):

$$ 0\le A-\frac{S\bar S}B-\frac{\bar SS}B+\frac{|S|^2}{B^2}\cdot B=A-\frac{|S|^2}B $$

Step 5 · Rearrange

$$ |S|^2\le AB \qquad\Longrightarrow\qquad \left|\sum_{k=1}^na_k\bar b_k\right|^2\le\left(\sum_{k=1}^n|a_k|^2\right)\left(\sum_{k=1}^n|b_k|^2\right) $$

Step 6 · Adjust from $\bar b_k$ to $b_k$

Since $|b_k|=|\bar b_k|$ for every $k$, replacing $b_k$ by $\bar b_k$ throughout Step 5 changes nothing on the right side, giving the stated form:

$$ \boxed{\left|\sum_{k=1}^na_kb_k\right|^2\le\left(\sum_{k=1}^n|a_k|^2\right)\left(\sum_{k=1}^n|b_k|^2\right)} $$

Equality condition

Equality holds exactly when $\displaystyle\sum_k|a_k-\lambda \bar b_k|^2=0$ for the optimal $\lambda$, i.e. when $\{a_k\}$ and $\{\bar b_k\}$ (equivalently $\{b_k\}$, up to conjugation convention) are proportional.

Verification, $n=2$

$a_1=1+i,\ a_2=2,\ b_1=1,\ b_2=1-i$: $\ |a_1b_1+a_2b_2|^2=|(1+i)+2(1-i)|^2=|3-i|^2=10$. Right side: $(|1+i|^2+|2|^2)(|1|^2+|1-i|^2)=(2+4)(1+2)=18$. Indeed $10\le18$ ✓.