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let $a_{i},b_{i}>0$, show that $$\sum_{i=1}^{n}a_{i}b_{i}\ge \dfrac{2}{n+\sqrt{\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}}}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}\tag{1}$$

I try:since $$\sum_{i=1}^{n}\dfrac{b_{i}}{a_{i}}\sum_{i=1}^{n}\dfrac{a_{i}}{b_{i}}\ge n^2$$ We just have to prove the inequality $$\sum_{i=1}^{n}a_{i}b_{i}\ge\dfrac{1}{n}\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}$$ This looks like a Chebyshev's inequality, but unfortunately the monotonicity of the two sequences is not clear, so it can not be applied directly，so How to prove inequality $(1)$

math110
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    Counter example to the second inequality : $n=2$ and $a_1=b_2=\varepsilon$, $a_2=b_1=1$, then the left hand side is $2\varepsilon$ while the right hand side is $\frac{(1-\varepsilon)^2}{2}$, rewriting your inequality gives $(1-\varepsilon)^2\leq 0$ which is obviously false when $\varepsilon\neq 1$. This example gives equality in the first inequality. – P. Quinton Apr 13 '20 at 08:26
  • @P.Quinton you can just have $\epsilon = 0$ and appeal to continuity – mathworker21 Apr 14 '20 at 06:25

1 Answers1

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Proof: We have the following identity: \begin{align} &\sum a_i b_i - \frac{2}{n + \sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}} \sum a_i \sum b_i\\ =\ & \sum \frac{b_i}{a_i} \left(a_i - \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}} {\left(n+\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}\right)\sum \frac{a_i}{b_i}}\sum a_i\right)^2. \end{align} We are done.

$\phantom{2}$

Remark: It is not hard to verify the identity above. Let $A = \sum \frac{a_i}{b_i}$, $B = \sum \frac{b_i}{a_i}$, $C = \sum a_i$ and $D = \sum b_i$. We have \begin{align} &\sum \frac{b_i}{a_i} \left(a_i - \frac{\frac{a_i}{b_i}\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}} + \sum \frac{a_i}{b_i}} {\left(n+\sqrt{\sum \tfrac{b_i}{a_i} \sum \tfrac{a_i}{b_i}}\right)\sum \frac{a_i}{b_i}}\sum a_i\right)^2\\ =\ & \sum \frac{b_i}{a_i} \left(a_i - \frac{\frac{a_i}{b_i}\sqrt{AB} + A}{(n+\sqrt{AB})A}C\right)^2\\ =\ & \sum \frac{b_i}{a_i} \left(a_i^2 - 2a_i \frac{\frac{a_i}{b_i}\sqrt{AB} + A}{(n+\sqrt{AB})A}C + \frac{(\frac{a_i}{b_i}\sqrt{AB} + A)^2}{(n+\sqrt{AB})^2A^2}C^2\right)\\ =\ & \sum \left(a_ib_i - \frac{2a_i\sqrt{AB} + 2b_iA}{(n+\sqrt{AB})A}C + \frac{\frac{a_i}{b_i}AB + 2A\sqrt{AB} + \frac{b_i}{a_i}A^2}{(n+\sqrt{AB})^2A^2}C^2 \right)\\ =\ & \sum a_ib_i - \frac{2C\sqrt{AB} + 2DA}{(n+\sqrt{AB})A}C + \frac{A\cdot AB + n\cdot 2A\sqrt{AB} + B\cdot A^2}{(n+\sqrt{AB})^2A^2}C^2\\ =\ &\sum a_ib_i - \frac{2}{n+\sqrt{AB}} CD. \end{align} We are done.

River Li
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