Proof of Cauchy-Schwarz by induction

Question

I have reached a blocking point on a problem I am working on.

Let $n \in \mathbb{N}$ be arbitrary and let $a_1,...,a_n, b_1,...,b_n\in \mathbb{R}$. Prove by induction on $n$ that $$(\sum_{i=1}^n a_ib_i)^2 \le(\sum_{i=1}^na_i^2)(\sum_{i=1}^nb_i^2) $$.

I have shown that in the base case, $$(a_1b_1)^2=a_1^2b_i^2$$

Now assuming the statement holds when $n=k\in\mathbb{N}$, I set up the following inequality.

$$(\sum_{i=1}^{n+1} a_ib_i)^2 \le(\sum_{i=1}^{n+1}a_i^2)(\sum_{i=1}^{n+1}b_i^2) $$

I will now show that this inequality holds using the induction hypothesis.

$$(a_{n+1}b_{n+1} + \sum_{i=1}^{n} a_ib_i)^2 \le(a_{n+1}^2 + \sum_{i=1}^{n}a_i^2)( b_{n+1}^2\sum_{i=1}^{n}b_i^2) $$

I then decide to expand each side giving

$$(a_{n+1}b_{n+1})^2 + 2a_ib_i\sum_{i=1}^{n} a_ib_i+(\sum_{i=1}^{n} a_ib_i)^2 \le (a_{n+1}b_{n+1})^2 + a_{n+1}^2 (\sum_{i=1}^{n}b_i^2)+ b_{n+1}^2 (\sum_{i=1}^{n}a_i^2) + (\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}a_i^2))$$

I subtract $(a_{n+1}b_{n+1})^2$ from both sides giving:

$$ 2a_ib_i\sum_{i=1}^{n} a_ib_i+(\sum_{i=1}^{n} a_ib_i)^2 \le a_{n+1}^2 (\sum_{i=1}^{n}b_i^2)+ b_{n+1}^2 (\sum_{i=1}^{n}a_i^2) + (\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}a_i^2))$$

I also notice the induction hypothesis but I am unsure if I am simply able to remove it and leave the following. Even if I am able to remove the induction hypothesis, where would I go from there?

$$ 2a_ib_i\sum_{i=1}^{n} a_ib_i \le a_{n+1}^2 (\sum_{i=1}^{n}b_i^2)+ b_{n+1}^2 (\sum_{i=1}^{n}a_i^2)$$

Answer 1

You wrote:

$$(\sum_{i=1}^{n+1} a_ib_i)^2 \le(\sum_{i=1}^{n+1}a_i^2)(\sum_{i=1}^{n+1}b_i^2) $$

I will now show that this inequality holds using the induction hypothesis.

$$(a_{n+1}b_{n+1} + \sum_{i=1}^{n} a_ib_i)^2 \le(a_{n+1}^2 + \sum_{i=1}^{n}a_i^2)( b_{n+1}^2\sum_{i=1}^{n}b_i^2) $$

The second equation here does not follow from the first by the induction hypothesis at all. What you COULD say, using the induction hypothesis is \begin{align} (\sum_{i=1}^{n+1} a_ib_i)^2 &= \left( a_{n+1}b_{n+1} + \sum_{i=1}^{n} a_ib_i \right)^2\\ &= \left( a_{n+1}b_{n+1} \right)^2 + 2a_{n+1}b_{n+1}\sum_{i=1}^{n} a_ib_i + \left(\sum_{i=1}^{n} a_ib_i \right)^2\\ &= \left( a_{n+1}b_{n+1} \right)^2 + 2a_{n+1}b_{n+1}\sum_{i=1}^{n} a_ib_i + \left(\sum_{i=1}^{n}a_i^2\sum_{i=1}^{n}b_i^2\right) \end{align} but whether that would be of any use to you is not clear to me. Indeed, it's not clear to me at all that induction is the way to approach this problem, but that's a separate matter. (The proof using the discriminant is my own favorite.)

Also: your base-case is wrong -- the subscripts "i" in there should be $1$s.