# $\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$ [closed]

Let $a_1, a_2,....,a_n, b_1, b_2,...,b_n$, let $\frac{b_1}{a_1} = max \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ , $\frac{b_n}{a_n} = min \{\frac{b_i}{a_i}, i=1,2, \cdots n \}$ show that:

$$\frac{1}{2}(\frac{b_1}{a_1}-\frac{b_n}{a_n})^2(\sum_{1}^{n}{a_i^2 }) ^2 \ge (\sum_{1}^{n}{a_i^2 }) (\sum_{1}^{n}{b_i^2 })-(\sum_{1}^{n}{a_ib_i })^2$$

## closed as off-topic by Semiclassical, Shailesh, Math1000, heropup, user223391 Aug 28 '16 at 13:53

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Semiclassical, Shailesh, Math1000, heropup, Community
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• This inequality is equivalent to $$\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 \geq \left(||B||\sin\theta\right)^2$$ Where $A=\left(a_1,a_2,\ldots,a_2\right)$, $B=\left(b_1,b_2,\ldots,b_2\right)$, and $\theta$ is the angle between $A$ and $B$. – Hrhm Aug 26 '16 at 17:13
• Could you show me detail? Please – Oai Thanh Đào Aug 26 '16 at 17:26
• Sure. We know that $$\sum_1^n a_ib_i=||A||||B||\cos\theta$$ and $$\sum_1^n a_i^2=||A||^2$$. I'm sure you can figure it out from there. This doesn't really help solve the inequality though, sorry about that. – Hrhm Aug 26 '16 at 17:29