Mathematics Olympic Question answer: Three kinds of solution of an inequality problem

Set $a _1, a_2, \cdots, a_n\in\mathbf{n^*}$, and different. Verification: $${a_1\over1^2} + {a_2\over2^2} + \cdot + {A_n\over n^2} \ge {1\over1} + {1\over2} + \cdots + {1\over n}.$$

Solution One:

Consider using the basic inequalities $a + b\ge 2\sqrt{ab}$ to eliminate the molecules on the left. $$\because {A_k\over 1^2} + {1\over a_k} \ge 2\cdot{1\over k},\ k=1, 2, \cdots n.$$ $$\therefore \sum_{k=1}^{n}{a_k\over k ^2} + \sum_{k=1}^{n}{1\over A_k} \ge 2\cdot\sum_{k=1}^{n}{1\over k}.$$ and $a _k\in\mathbf{n^*}$ and not equal, so $$\sum_{k=1}^{n}{ 1\over A_k} \le \sum_{k=1}^{n}{1\over k}.$$ Note there is not necessarily $a_k \ge k$.

From this you can get $$\sum_{k=1}^{n}{a_k\over k^2} \ge \sum_{k=1}^{n}{1\over k}.$$

Solution Two:

We can also try to use the Cauchy inequality $$\left (\sum_{i = 1}^{n}a_ib_i\right) ^2 \le \left (\sum_{i = 1}^{n}a_i^2\right) \left (\sum_{i = 1}^{n} B_i^2\right) $$ Proof: $$\left (1 + {1\over2} + \cdots + {1\over n}\right) ^2 = \left (\sum_{k = 1}^{n}{1\over k}\right) ^2 = \le ft (\sum_{k = 1}^{n}{\sqrt{a_k} \over k}\cdot {1\over \sqrt{a_k}}\right) ^2 \le \sum_{k = 1}^{n}{a_k \over k^2} \cdot \sum_{ K = 1}^{n}{1\over a_k}$$ and $a _k\in\mathbf{n^*}$ and are not equal, so $$\sum_{k=1}^{n}{1\over a_k} \le \sum_{k=1}^{n}{1\over k}.$$ thus To get $$\sum_{k=1}^{n}{a_k\over k^2} \ge \sum_{k=1}^{n}{1\over k}.$$

Solution Three:

The subject can also be proved by the sort inequalities.

Set $b _1, b_2, \cdots, b_n$ is the rearrangement of $a _1, a_2, \cdots, a_n$, and satisfies: $ $b _1 < b_2 < \cdots b_n.$$ and $$1 > {1\over2^2} &G T {1\over3^2} > \cdots > {1\over n^2}.$$ so (disorderly and $\ge$ reverse order and): $ $a _1 + {a_2 \over 2^2} + \cdots + {a_n \over n^2} \ge b_1 + {b_2 \over 2^2} + \cdots + {b_n \over n^2}.$$ because $b _1, b_2, \cdots, b_n$ are non-equal positive integers, $ $b _1 \ge 1,\ b_2\ge2,\ \cdots,\ B_n \ge n.$$ thereby $$1 + {1\over 2} + \cdots + {1\over n} \le b_1 + {b_2 \over 2^2} + \cdots + {b_n \over n^2} \le a_1 + {a_2 \over 2^2} + \cdots + {A_n \over n^2}.$$

Commentary:

The solution of the first method, two similar, both use the identity $${1\over k} = {\sqrt{a_k} \over k} \cdot {1 \over \sqrt{a_k}}.$$ solution Three uses the sort inequality to obtain the result quickly, is three kind of practice the most concise.

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Mathematics Olympic Question answer: Three kinds of solution of an inequality problem