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$(0 < m\le a_k/b_k \le M)\implies |A||B|\le{\small\dfrac{m + M}{2\sqrt{mM}}}(A\cdot B)$ - Printable Version

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$(0 < m\le a_k/b_k \le M)\implies |A||B|\le{\small\dfrac{m + M}{2\sqrt{mM}}}(A\cdot B)$ - elim - 12-15-2012 04:45 AM


Let $m,\, M,\, a_k,\, b_k \in \mathbb{R}$ such that $\displaystyle{a_k > 0, \, b_k > 0,\, 0 < m \le \frac{a_k}{b_k} \le M\; (k = 1,\ldots , n)}$

Show that $\small\displaystyle{\big(\sum_{k=1}^n a_k^2\big)^{1/2}\big(\sum_{k=1}^n b_k^2\big)^{1/2} \le \frac{m + M}{2\sqrt{m M}} \sum_{k=1}^n a_k b_k}$


RE:$(0 < m\le a_k/b_k \le M)\implies (|A||B|)\le \frac{m + M}{2\sqrt{mM}}(A\cdot B)$ - elim - 12-15-2012 05:30 AM


$\displaystyle{\left(m \le \frac{a_k}{b_k} \le M \right) \Rightarrow \left(\left(\frac{a_k}{b_k} -m\right)\left(M- \frac{a_k}{b_k}\right) \ge 0\right) \Rightarrow \left((m+M) a_k b_k \ge a_k^2 + mM b_k^2 \right )}$

$\displaystyle{ (m+M)\sum_{k=1}^n a_k b_k \ge \sum_{k=1}^n a_k^2 +mM \sum_{k=1}^n b_k^2 \overset{AM\ge GM}{\ge} 2\sqrt{mM}\left(\sum_{k=1}^n a_k^2 \right )^{1/2} \left(\sum_{k=1}^n b_k^2 \right )^{1/2}} $