Proving Cauchy-Schwarz related proof using induction So the first thing I was asked to prove was this: 
If $a_1,a_2,...,a_n$ and $b_a,b_2,...,b_n$ are real numbers, use induction to show. $$\bigg(\sum_{k=1}^na_k^2\bigg)\bigg(\sum_{k=1}^nb_k^2\bigg)-\bigg(\sum_{k=1}^na_kb_k\bigg)^2=\sum_{1\ \le\  i\lt \ j \le\ n}^n(a_ib_j-a_ib_j)^2$$ I'm having trouble dealing with the indices for the summation on the right side of the equation.
I know how to do induction and I realize that I can't start my base case with $1$. I'd have to start it with $2$, but it's the next cases that start to give me trouble and start to confuse me. Any help would be much appreciated!
The second thing it wants us to do is to prove $$\bigg(\sum_{k=1}^na_k^2\bigg)\bigg(\sum_{k=1}^nb_k^2\bigg)\ge\bigg(\sum_{k=1}^na_kb_k\bigg)^2$$ Using the proof of the above equation. I'm hoping though once I understand the indices from the above question I'll be able to prove the second one.
EDIT: (MY WORK)
Consider the base case $P(2):$ $$\bigg(\sum_{k=1}^2a_k^2\bigg)\bigg(\sum_{k=1}^2b_k^2\bigg)-\bigg(\sum_{k=1}^2a_kb_k\bigg)^2=\sum_{1\ \le\  i\lt \ j \le\ 2}^2(a_ib_j-a_ib_j)^2$$
Which becomes: $$(a_1^2+a_1^2)(b_1^2+b_1^2)-(a_1b_1+a_2b_2)^2=(a_1b_2-a_2b_1)^2$$
Which simplifies into: $$(a_1b_1)^2+(a_1b_2)^2+(a_2b_1)^2+(a_2b_2)^2-(a_1b_1)^2-(a_2b_2)^2-2(a_2b_2)^2(a_1b_1)^2=(a_1b_2)^2+(a_2b_1)^2-2(a_2b_1)^2(a_1b_2)^2$$
Canceling like terms yields: $$(a_2b_2)^2(a_1b_1)^2=(a_2b_1)^2(a_1b_2)^2$$
EDIT $2$:
$$(a_2b_2)^2(a_1b_1)^2=(a_2b_1)^2(a_1b_2)^2\rightarrow (a_2b_2a_1b_1)^2=(a_2b_1a_1b_2)^2$$
Now we can use the fact that multiplication is commutative to say that: $$(a_2b_2a_1b_1)^2=(a_2b_2a_1b_1)^2$$
 A: Write things out carefully. For $n + 1$, want to show
$$\left(\sum_{k=1}^na_k^2 + a_{n+1}^2\right)\left(\sum_{k=1}^nb_k^2 +b_{n+1}^2\right)-\left(\sum_{k=1}^na_kb_k + a_{n+1}b_{n+1}\right)^2\\
=\sum_{1\ \le\  i\le \ j \le\ n}^n(a_ib_j-a_jb_i)^2 + \sum_{i=1}^n (a_ib_{n+1} - a_{n+1}b_i)^2$$
Distribute to separate out the terms you already have control of:
$$\left(\sum_{k=1}^na_k^2\right)\left(\sum_{k=1}^nb_k^2\right) + a_{n+1}^2\sum_{k=1}^nb_k^2 + b_{n+1}^2\sum_{k=1}^na_k^2 + \left(a_{n+1}b_{n+1}\right)^2\\ -\left(\sum_{k=1}^na_kb_k\right)^2 - 2a_{n+1}b_{n+1}\left(\sum_{k=1}^na_kb_k\right) - \left(a_{n+1}b_{n+1}\right)^2\\
$$
Two terms cancel, and then use induction hypothesis:
$$= \sum_{1\ \le\  i\le \ j \le\ n}^n(a_ib_j-a_jb_i)^2 + a_{n+1}^2\sum_{k=1}^nb_k^2 + b_{n+1}^2\sum_{k=1}^na_k^2\\ - 2a_{n+1}b_{n+1}\left(\sum_{k=1}^na_kb_k\right)\\
$$
it isn't too hard to show this is equal to the RHS of the desired expression.
The second thing is a one-liner once you establish the first fact; notice the two terms being compared in the second expression are the LHS of the first expression. Do you think the RHS is $\geq 0$? That is equivalent to the second expression.
