Inequality proof 2

Question

How to prove the inequality : for real numbers $\alpha_1, \ldots \alpha_n, \beta_1, \ldots \beta_n$:

$$\sqrt{(\alpha_1 + \beta_1)^2+\cdots+(\alpha_n + \beta_n)^2} \leq \sqrt{\alpha_1^2 + \cdots + \alpha_n^2}+\sqrt{\beta_1^2+\cdots+\beta_n^2}.$$

Thanks!

score 3 · Answer 1 · answered May 20 '13 at 20:47

3

One easy way is to square both sides. Then use Cauchy-Schwarz to show the inequailty between the terms that don't vanish on both sides.

answered May 20 '13 at 20:47

sigmatau

2,622

score 0 · Accepted Answer · answered May 20 '13 at 20:47

Hint: Square both sides

$(\alpha_1 + \beta_1)^2+\cdots+(\alpha_n + \beta_n)^2 \leq \alpha_1^2 + \cdots + \alpha_n^2+\beta_1^2+\cdots+\beta_n^2+ 2\sqrt{\left(\alpha_1^2 + \cdots + \alpha_n^2\right) \left(\beta_1^2+\cdots+\beta_n^2\right)}$

$\alpha_1 \beta_1+\cdots+\alpha_n \beta_n \leq \sqrt{\left(\alpha_1^2 + \cdots + \alpha_n^2\right) \left(\beta_1^2+\cdots+\beta_n^2\right)}$

Inequality proof 2

2 Answers2