<s>[INST] Mathematical Induction Proof for $(\sum_{k=1}^{n}a^2_k)(\sum_{k=1}^{n}b^2_k) \geq (\sum_{k=1}^{n}a_kb_k)^2$ Mathematical Induction Proof $$(\sum_{k=1}^{n}a^2_k)(\sum_{k=1}^{n}b^2_k) \geq (\sum_{k=1}^{n}a_kb_k)^2$$
Base case: $n = 2$, $$(a^2_1+a^2_2)(b^2_1+b^2_2) \geq (a_1b_1 
+ a_2b_2)^2$$
$$(a^2_1+a^2_2)(b^2_1+b^2_2) - (a_1b_1 
+ a_2b_2)^2 \geq 0$$
$$(a_1b_2-a_2b_1)\geq 0$$
Let $$\sum_{i=1}^{k}a^2_i = A^2, \sum_{i=1}^{k}b^2_i = B^2,$$
$$(A^2 + a^2_{k+1})(B^2 + b^2_{k+1}) =$$
$$(A^2B^2 + a^2_{k+1}b^2_{k+1} + A^2b^2_{k+1} + B^2a^2_{k+1})$$
This step is where I don't understand:
$$(A^2B^2 + a^2_{k+1}b^2_{k+1} + A^2b^2_{k+1} + B^2a^2_{k+1}) \geq (AB + a_{k+1}b_{k+1})^2 \geq (\sum_{k=1}^{n}a_kb_k)^2 $$
I know that $$(AB + a_{k+1}b_{k+1})^2 =  A^2B^2 + a^2_{k+1}b^2_{k+1} + 2ABa_{k+1} b_{k+1}$$
and that $$A^2b^2_{k+1} + B^2a^2_{k+1} \geq 0$$
hence,
$$A^2b^2_{k+1} + B^2a^2_{k+1} \geq 2ABa_{k+1} b_{k+1}$$
hence, $$(A^2B^2 + a^2_{k+1}b^2_{k+1} + B^2a^2_{k+1} + A^2b^2_{k+1}) \geq (AB + a_{k+1}b_{k+1})^2 $$
but why $$(AB + a_{k+1}b_{k+1})^2 \geq (\sum_{k=1}^{n}a_kb_k)^2 $$?
 [/INST] By the assumption of the induction twice and by AM-GM we obtain:
$$\sum_{i=1}^{n+1}a_i^2\sum_{i=1}^{n+1}b_i^2=\sum_{i=1}^{n}a_i^2\sum_{i=1}^{n}b_i^2+a_{n+1}^2\sum_{i=1}^nb_i^2+b_{n+1}^2\sum_{i=1}^na_i^2+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2\sqrt{a_{n+1}^2b_{n+1}^2\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2}+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2a_{n+1}b_{n+1}\sum_{i=1}^na_ib_i+a_{n+1}^2b_{n+1}^2=\left(\sum_{i=1}^{n+1}a_ib_i\right)^2.$$
</s>