Mathematical Olympiad: (Russia,1992)

Question

Problem

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.

Answer 1

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.

Answer 2

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.