Prove Cauchy's inequality: $(\sum_{i=1}^n a_i b_i)^2 \le \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2.$

Question

Prove that for any real number $a_i, b_i, i = 1,2,...,n$: $$\left(\sum_{i=1}^n a_i b_i\right)^2 \le \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2.$$ Proof in my book goes: $$\sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2-\left(\sum_{i=1}^n a_i b_i\right)^2 = \sum_{i,j=1}^na_i^2b_j^2-\sum_{i,j=1}^na_ib_ia_jb_j =\frac{1}{2}\sum_{i,j=1}^n(a_ib_j-b_ia_j)^2\ge0$$ But I can't see why there is $\frac{1}{2}$ and I would be glad if someone explained it. Tia.

Answer 1

Take the last term of the equality and expand it:

$$\sum_{i,j=1}^n(a_ib_j-b_ia_j)^2 = \sum_{i,j = 1\dots n} a_i^2b_j^2 + b_i^2a_j^2 - 2a_ib_jb_ia_j = \sum_{i,j = 1\dots n} 2a_i^2b_j^2 - 2a_ib_jb_ia_j $$

And here you can see that this is exactly the double of the term you had before. So in order to get an equality you have to take an $\frac{1}{2}$ factor.

EDIT

Please note that the last passage I did above is possible since we can interchange the indexes. To be more clear:

$$\sum_{i,j = 1\dots n} a_i^2b_j^2 + b_i^2a_j^2 - 2a_ib_jb_ia_j = \\ \sum_{i = 1\dots n}\sum_{j = 1\dots n} \left(a_i^2b_j^2 - a_ib_jb_ia_j\right) + \sum_{i = 1\dots n}\sum_{j = 1\dots n} \left(a_j^2b_i^2 - a_ib_jb_ia_j\right)$$

Now you can swap $i$ with $j$ in the last double sum and you can conclude.