Sum of an arithmetic series

Question

Let $a_1, a_2, ..., a_{n-1}$ is strictly positive, not equal, real numbers

If $a_1 + a_2 + ... + a_{n-1}$ = $n-1$. Then what will be the answer for

$\frac{n}{\frac{1}{a_1} + \frac{1}{a_2}+ ... + \frac{1}{a_{n-1}}}$

I know it's an elementary but a bit difficult for me to catch it

Answer 1

Note that for $a_i\ge 0$ by HM-AM inequality we have that

$$\frac{n-1}{\frac{1}{a_1} + \frac{1}{a_2}+ ... + \frac{1}{a_{n-1}}}\le \frac{a_1 + a_2 + ... + a_{n-1} }{n-1}=1$$

then we can bound the expression by

$$\frac{n}{\frac{1}{a_1} + \frac{1}{a_2}+ ... + \frac{1}{a_{n-1}}}\le \frac{n}{n-1}$$

In general we can't find a closed expression for its value.

Answer 2

With $n = 2$, $a_1+a_2 = 1$. If $a_1 = x, a_2 = 1-x$, then: $$\frac{2}{\frac{1}{x} + \frac{1}{1-x}} = \frac{2x(1-x)}{(1-x)+x} = 2x(1-x) = -2x^2+2x.$$

The maximum has x-coordinate $-\frac{b}{2a}$ or $-\frac{2}{2\cdot -2} = \frac{1}{2}$. Therefore, the maximum value is $-2 \cdot (\frac{1}{2})^2 + 2(\frac{1}{2}) = -\frac{1}{2} + 1 = \frac{1}{2}$. Therefore, the fraction is always less than $1$ for $n=2$.