# Prove Cauchy–Schwarz inequality by Induction

I got stuck while proving the Cauchy–Schwarz inequality:

$$\bigg(\sum_{i=1}^{n}a_{i}^2\bigg).\bigg(\sum_{i=1}^{n}b_{i}^2\bigg)\ge \bigg(\sum_{i=1}^{n}a_ib_i\bigg)^2$$

My Attempt:

For $n=1$ it is trivial. So we might have to try for $n=2$, which gives $$P(2): (a_1^2+a_2^2)(b_1^2++b_2^2)\ge (a_1b_1+a_2b_2)^2\\a_1^2b_1^2+a_1^2b_2^2+a_2^2b_1^2+a_2^2b_2^2\ge a_1^2b_1^2+2a_1b_1a_2b_2+a_2^2b_2^2\\a_1^2b_2^2-2a_1b_2a_2b_1+a_2^2b_1^2=(a_1b_2-a_2b_1)\ge 0,$$ which is true.

Now assuming the inequality is true for any $k$ terms. $P(k):\bigg(\sum_{i=1}^{k}a_i^2\bigg)\bigg(\sum_{i=1}^{k}b_i^2\bigg)\ge\bigg(\sum_{i=1}^{k}a_ib_i\bigg)^2$

For $k+1$ terms, $$\bigg(\sum_{i=1}^{k+1}a_i^2\bigg)\bigg(\sum_{i=1}^{k+1}b_i^2\bigg)=\bigg(\sum_{i=1}^{k}a_i^2+a_{k+1}^2\bigg)\bigg(\sum_{i=1}^{k}b_i^2+b_{k+1}^2\bigg)\\=\sum_{i=1}^{k}a_i^2\sum_{i=1}^{k}b_i^2+b_{k+1}^2\sum_{i=1}^{k}a_i^2+a_{k}^2\sum_{i=1}^{k}b_i^2+a_{k+1}^2b_{k+1}^2$$ Since $$\bigg(\sum_{i=1}^{k+1}a_ib_i\bigg)^2=\bigg(\sum_{i=1}^ka_ib_i+a_{k+1}b_{k+1}\bigg)^2=\bigg(\sum_{i=1}^ka_ib_i\bigg)^2+2a_{k+1}b_{k+1}\sum_{i=1}^ka_ib_i+a_{k+1}^2b_{k+1}^2,$$ for P(k+1) to be true we need to prove that $$b_{k+1}^2\sum_{i=1}^{k}a_i^2+a_{k+1}^2\sum_{i=1}^{k}b_i^2\ge 2a_{k+1}b_{k+1}\sum_{i=1}^ka_ib_i\\\implies a_{k+1}^2\sum_{i=1}^{k}b_i^2-2a_{k+1}b_{k+1}\sum_{i=1}^ka_ib_i+b_{k+1}^2\sum_{i=1}^{k}a_i^2\ge 0.$$

How do I prove the last step ? I got a hint from link relating it as a bivariate polynomial, but I absolutely have no idea how to deal with it. or is there a better way to prove the last statement ?

Note: I tried to show that the LHS is greater than a perfect square by setting $a_i=\tfrac{a_{k+1}}{\sqrt{k}}$, from $P(k)$ we get $a^2_{k+1}\sum_{i=1}^k b_i^2\ge \tfrac{a^2_{k+1}}{k}(\sum_{i=1}^{k}a_i)^2$ and similarly by setting $b_i=\tfrac{b_{k+1}}{\sqrt{k}}$ we get $b^2_{k+1}\sum_{i=1}^k a_i^2\ge \tfrac{b^2_{k+1}}{k}(\sum_{i=1}^{k}b_i)^2$. But again I got tuck at proving $\sum_{i=1}^ka_ib_i\le \tfrac{\sum a_i\sum b_i}{k}$. Is it wrong to approach like this ?

## 2 Answers

$$a_{k\color{red}{+1}}^2\sum_{i=1}^{k}b_i^2-2a_{k+1}b_{k+1}\sum_{i=1}^ka_ib_i+b_{k+1}^2\sum_{i=1}^{k}a_i^2\ge 0$$

Hint:  after correcting the index of $a_{k\color{red}{+1}}^2\,$, consider the LHS of the above as a quadratic in $a_{k+1}\,$. Its reduced discriminant is:

$$\frac{1}{4}\Delta = b_{k+1}^2 \left(\sum_{i=1}^ka_ib_i\right)^2 - b_{k+1}^2 \left(\sum_{i=1}^{k}a_i^2\right)\left(\sum_{i=1}^{k}b_i^2\right) \le 0$$

• tnx. pls check my edit. could you just comment on my approach to prove the last statement ? – ss1729 Jul 21 '17 at 18:12
• ohh i think i got into trouble thinking about the bivariate polynomial and all. As you said if i consider it as a polynomial with variable $a_{k+1}$, it is the equation of an up parabola. Thus got either one or no real solutions, ie. discriminant, $B^2-4AC\le 0$, which can be proved from $P(k)$. is that correct ? – ss1729 Jul 21 '17 at 18:18
• @ss1729 The last comment is correct. A quadratic with non-positive discriminant does not change sign. Since the leading coefficient is positive, it means the quadratic is non-negative, which is precisely what needed to be proved. – dxiv Jul 21 '17 at 18:21
• thnx. i think tht complete the proof – ss1729 Jul 21 '17 at 18:27

Just $$\sum_{i=1}^{n}a_{i}^2\sum_{i=1}^{n}b_{i}^2- \bigg(\sum_{i=1}^{n}a_ib_i\bigg)^2=\sum_{1\leq i<j\leq n}(a_ib_j-a_jb_i)^2\geq0.$$ Another way.

Let $\vec{a}(a_1,a_,...,a_n)$ and $\vec{b}(b_1,b_2,...,b_n)$.

Thus, $$|\vec{a}\cdot\vec{b}|\leq|\vec{a}|\cdot|\vec{b}|$$ and we are done!

Another way.

For all real $x$ we have $$\sum_{i=1}^n(a_i+xb_i)^2\geq0$$ or $$\sum_{i=1}^nb_ix^2+2\sum_{i=1}^na_ib_ix+\sum_{cyc}a_i^2\geq0,$$ for which we need $$\left(\sum_{i=1}^na_ib_i\right)^2-\sum_{cyc}a_i^2\sum_{cyc}b_i^2\leq0$$ and we are done!