How to that prove this using induction?

Question

I have to prove that using induction:

Considering n $\in \mathbb N*$ and $x, y, \alpha_1, \alpha_2,...,\alpha_n, \beta_1, \beta_2,..., \beta_n$ real numbers then

$2xy(\alpha_1\beta_1 + \alpha_2\beta_2 + ... + \alpha_n\beta_n) \le x^2(\alpha_1^2 + \alpha_2^2 + ... + \alpha_n^2) + y^2(\beta_1^2 + \beta_1^2 + ... + \beta_n^2)$

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That's what I thought to do:

$2xy(\alpha_1\beta_1 + \alpha_2\beta_2 + ... + \alpha_n\beta_n) \le x^2(\alpha_1^2 + \alpha_2^2 + ... + \alpha_n^2) + y^2(\beta_1^2 + \beta_1^2 + ... + \beta_n^2) \equiv \sum_{i=1}^n (x\alpha_i - y\beta_i)^2 \ge 0$

Using the expression found:

Base case(1) : $\sum_{i=1}^n (x\alpha_i - y\beta_i)^2 = (x\alpha_1 - y\beta_1)^2 \ge 0$

Induction Step:

Assuming $\sum_{i=1}^k (x\alpha_i - y\beta_i)^2 \ge 0$ true, then we have to prove $\sum_{i=1}^{k+1} (x\alpha_i - y\beta_i)^2 \ge 0$.

$\sum_{i=1}^k (x\alpha_i - y\beta_i)^2 \ge 0$

$\sum_{i=1}^k (x\alpha_i - y\beta_i)^2 + (x\alpha_{k+1} - y\beta_{k+1})^2 \ge (x\alpha_{k+1} - y\beta_{k+1})^2$

$\sum_{i=1}^{k+1} (x\alpha_i - y\beta_i)^2 \ge (x\alpha_{k+1} - y\beta_{k+1})^2$

But,

$(x\alpha_{k+1} - y\beta_{k+1})^2 \ge 0$

Therefore,

$\sum_{i=1}^{k+1} (x\alpha_i - y\beta_i)^2 \ge 0$

Did I something wrong?

Answer 1

Everything is fine, perhaps instead of assuming the base case you could simply prove it for $k=1$

Answer 2

In the first line you write an equivalent form "sum of squares is nonnegative". That is obviously true so the proof can end there. Why bother with all the rest?