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 $$(AB + a_{k+1}b_{k+1})^2 \geq (\sum_{k=1}^{n}a_kb_k)^2 $$?
 A: There appears to be some confusion between the use of $k$ as a summation variable and an induction variable, among other issues such as using $n$ later where it's actually $k + 1$. First, rewrite what you're trying to prove by replacing $k$ with $i$ to get
$$\left(\sum_{i=1}^{n}a_i^2\right)\left(\sum_{i=1}^{n}b_i^2\right) \geq \left(\sum_{i=1}^{n}a_i b_i\right)^2 \tag{1}\label{eq1A}$$
Next, rewrite the part which you're unsure of by replacing $n$ on the right side by $k + 1$, giving
$$(AB + a_{k+1}b_{k+1})^2 \geq \left(\sum_{i=1}^{k+1}a_i b_i\right)^2 = \left(\sum_{i=1}^{k}a_i b_i + a_{k+1}b_{k+1}\right)^2 \tag{2}\label{eq2A}$$
The induction hypothesis step is that \eqref{eq1A} is true for $n = k$, i.e.,
$$\left(\sum_{i=1}^{k}a_i^2\right)\left(\sum_{i=1}^{k}b_i^2\right) \geq \left(\sum_{i=1}^{k}a_i b_i\right)^2 \tag{3}\label{eq3A}$$
The stated definitions you're using are
$$\sum_{i=1}^{k}a^2_i = A^2, \; \sum_{i=1}^{k}b^2_i = B^2 \tag{4}\label{eq4A}$$
Using the unstated requirement that all of of the values are non-negative, then by substituting the values from \eqref{eq4A} into \eqref{eq3A}, and using $A^2B^2 = (AB)^2$, gives
$$(AB)^2 \ge \left(\sum_{i=1}^{k}a_i b_i\right)^2 \implies AB \ge \sum_{i=1}^{k}a_i b_i \tag{5}\label{eq5A}$$
Thus, adding $a_{k+1}b_{k+1}$ to both sides and squaring gives \eqref{eq2A}.
Using the other parts of your proof gives that \eqref{eq1A} is also true for $n = k + 1$, so from \eqref{eq1A} being true for the base case of $n = 1$ since $a_1b_2 \ge a_1b_1$, and using your extra base case of $n = 2$, gives that, by induction, \eqref{eq1A} is true for all $n \ge 1$.
A: By the assumption of the induction twice and by AM-GM we obtain:
$$\sum_{i=1}^{n+1}a_i^2\sum_{i=1}^{n+1}b_i^2=\sum_{i=1}^{n}a_i^2\sum_{i=1}^{n}b_i^2+a_{n+1}^2\sum_{i=1}^nb_i^2+b_{n+1}^2\sum_{i=1}^na_i^2+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2\sqrt{a_{n+1}^2b_{n+1}^2\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2}+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2a_{n+1}b_{n+1}\sum_{i=1}^na_ib_i+a_{n+1}^2b_{n+1}^2=\left(\sum_{i=1}^{n+1}a_ib_i\right)^2.$$
