Problem of the Week

Problem 9 solution

Congratulations to Ivan at St Bede’s Inter-church School and Vedant for solving Problem 9.

To approach this problem, let’s start with a pair of positive real numbers $a$ and $b$ .

After some experimentation, it becomes apparent that for any positive real number $x$ , $x + \frac{1}{x} \geq 2$ . However, we must also prove this.

Consider the quadratic $(x-1)^2$ , where $x$ is a positive real number. For all real values of $x$ , $(x-1)^2 \geq 0$ . Consequently,

$x^2 - 2x + 1 \geq 0.$

Since $x$ is positive, we may divide by $x$ without changing the inequality. Therefore,

$x + \frac{1}{x} \geq 2.$

Now, applying this to the general case, we know that

$n_1 + \frac{1}{n_1} + n_2 + \frac{1}{n_2} + \ldots + n_m + \frac{1}{n_m} \geq 2m,$

by applying our earlier proof to all values of $n_i$ .

If the sum of $n_1 + n_2 + \ldots + n_m$ does not exceed $S$ , we may also say

$n_1 + n_2 + \ldots + n_m \geq 2m - S.$

Thus, the inequality is proven.

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