Prove that: $2pq\sum_{k=1}^{n} a_kb_k\leq p^2\sum_{k=1}^{n} {a_k}^{2}+q^2 \sum_{k=1}^{n} {b_k}^{2}$

Question

$\forall n\in\mathbb{N},\forall p,q,a_i,b_i\in\mathbb{R}$ prove that: $$2pq\sum_{k=1}^{n} a_kb_k\leq p^2\sum_{k=1}^{n} {a_k}^{2}+q^2 \sum_{k=1}^{n} {b_k}^{2}.$$

It is like Cauchy-Scharwz inequality.

When $n=1$ ok.

Answer 1

By AM-GM and C-S we obtain: $$p^2\sum_{k=1}^{n} {a_k}^{2}+q^2 \sum_{k=1}^{n} {b_k}^{2}\geq2|pq|\sqrt{\sum_{k=1}^{n} {a_k}^{2}\sum_{k=1}^{n} {b_k}^{2}}\geq$$ $$\geq2|pq|\sqrt{\left(\sum_{k=1}^{n} a_kb_k\right)^2}=2|pq\sum_{k=1}^{n} a_kb_k|\geq2pq\sum_{k=1}^{n} a_kb_k.$$ Done!

Answer 2

Use the inequality $2(pa_k)(qb_k) \leq (pa_k)^2+(qb_k)^2$ and sum over $k$.

Answer 3

It is equivalent with: $$\sum_{k=1}^n (pa_k-qb_k)^2 \ge 0.$$