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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:

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

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

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    $\begingroup$ 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 $B$. $\endgroup$ – Hrhm Aug 26 '16 at 17:13
  • $\begingroup$ Could you show me detail? Please $\endgroup$ – Oai Thanh Đào Aug 26 '16 at 17:26
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    $\begingroup$ 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. $\endgroup$ – Hrhm Aug 26 '16 at 17:29
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Note that \begin{align*} &\ \sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2 - \left(\sum_{i=1}^na_ib_i \right)^2-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2\left(\sum_{i=1}^na_i^2 \right)^2\\ =&\ \sum_{i,j=1}^n\left[a_i^2b_j^2- a_ib_ia_jb_j-\frac{1}{2}\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\ =&\ \frac{1}{2}\sum_{i,j=1}^n\left[a_i^2b_j^2 + a_j^2b_i^2- 2a_ib_ia_jb_j-\left(\frac{b_1}{a_1}-\frac{b_n}{a_n}\right)^2 a_i^2a_j^2\right]\\ \le&\ \frac{1}{2}\sum_{i,j=1}^n\left[a_i^2b_j^2 + a_j^2b_i^2- 2a_ib_ia_jb_j-\left(\frac{b_i}{a_i}-\frac{b_j}{a_j}\right)^2 a_i^2a_j^2\right]\\ =&\ 0. \end{align*} The inequality follows immediately.

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