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Problem

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My attempt

From AM-GM inequality it can be shown that $$\sum_{i=1}^{n}(a_i+b_i)^2\geq4\sum_{i=1}^{n}a_ib_i$$ Therefore, we have: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}a_ib_i$$ Now, on expanding the RHS, we get: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\left({1\over a_1b_1}+{1\over a_2b_2}+.....{1\over a_nb_n}\right)\left(a_1b_1+a_2b_2+....a_nb_n\right)$$ $$=4(n+\left({a_2b_2\over a_1b_1}+{a_1b_1\over a_2b_2}+....{a_nb_n\over a_1b_1}+{a_1b_1\over a_nb_n}\right)+\left({a_3b_3\over a_2b_2}+{a_2b_2\over a_3b_3}+....{a_nb_n\over a_2b_2}+{a_2b_2\over a_nb_n}\right)+...+\left({a_nb_n\over a_{n-1}b_{n-1}}+{a_{n-1}b_{n-1}\over a_nb_n}\right))$$

Using the fact that $a+{1\over a}\geq2$ we get: $$\sum_{i=1}^{n}{1\over a_ib_i}\sum_{i=1}^{n}(a_i+b_i)^2\geq4\left({1\over a_1b_1}+{1\over a_2b_2}+.....{1\over a_nb_n}\right)\left(a_1b_1+a_2b_2+....a_nb_n\right)\geq4(n+2(n-1+n-2+....+1))=4n^2$$

Is this proof correct? I am asking this because the proof given by the author of my textbook is quite different from this one.

Edit: Author's solution.

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  • $\begingroup$ I really like your proof, and I think it's fine. What does the textbook have to say on this question? $\endgroup$ – астон вілла олоф мэллбэрг Oct 7 '16 at 12:34
  • $\begingroup$ I've made an edit. Please check. $\endgroup$ – nls Oct 7 '16 at 12:44
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    $\begingroup$ I see. His proof is very nice, but his revolves around the Chebyshev (or Cauchy Schwarz or something like that ) inequality (the second part), while yours revolves around a simpler inequality of sum of number and reciprocal being greater than $2$. Your proof I have double checked, you should be good to go. $\endgroup$ – астон вілла олоф мэллбэрг Oct 7 '16 at 12:48
  • $\begingroup$ Thank you for your prompt response. $\endgroup$ – nls Oct 7 '16 at 12:49
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This is a correct proof, and is how I would have solved the problem. Problems like this usually have a large number of proofs, sometimes extremely different ones, and I wouldn't let the fact that this proof isn't in the book deter you. It would be useful though to learn the book's technique as well.

For the author's proof, the first equation given follows from $$\frac{1}{ab}\geq\frac{4}{(a+b)^2}$$ by simply adding a bunch of terms of that form, but the second equation. The second equation follows from Cauchy-Schwarz, which says (when restricted to real numbers) that $$\left(\sum_{i\in I} \alpha_i \beta_i\right)^2 \leq\sum_{j\in I} \alpha_j^2\sum_{k\in I} \beta_k^2$$ Applying this inequality with $\alpha_i=(a_i+b_i)^2$ and $\beta_i=(a_i+b_i)^{-2}$ produces the desired result.

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  • $\begingroup$ @Shrey Aryan I added an explication of the author's proof. $\endgroup$ – Stella Biderman Oct 7 '16 at 13:57
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Set $c_i = (a_i +b_i)^2$. Then by AM-GM, $a_ib_i \le c_i/4$. Therefore, our expression is bounded below by

$$4 \sum_{i=1}^n \frac{1}{c_i} \sum_{j=1}^n c_i =4 \sum_{i=1}^n c_i \times \sum_{i=1}^n \frac{1}{c_i}.$$

As for the RHS:

  1. By the harmonic - arithmetic mean inequality, $$\frac{n}{ \sum_{i=1}^n \frac{1}{c_i}} \le \frac{ {\sum_{i=1}^n c_i }}{n},$$ so result follows.
  2. Alternatively, use Cauchy Schwarz: $$\sum_{i=1}^n c_i \times \sum_{i=1}^n \frac{1}{c_i} \ge \left (\sum_{i=1}^n \sqrt{c_i} \frac{1}{\sqrt{c_i}}\right)^2 = n^2,$$ so result follows.
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