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 ?