- Thread starter
- #1

#### lfdahl

##### Well-known member

- Nov 26, 2013

- 719

where $n_1,n_2$ and $n_3$ are different positive integers, that satisfy:

$\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3} < 1.$

- Thread starter lfdahl
- Start date

- Thread starter
- #1

- Nov 26, 2013

- 719

where $n_1,n_2$ and $n_3$ are different positive integers, that satisfy:

$\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3} < 1.$

- Jan 30, 2018

- 378

+ \frac{1}{x_2}+ \frac{1}{x_3}\right)$ so we can treat $\frac{1}{x_1}

+ \frac{1}{x_2}+ \frac{1}{x_3}$ as a single quantity.

1- 3/x= (x- 3)/x is smallest when x= 3. Since $x_1$, $x_2$, and $x_3$ have be distinct integers, $1- \frac{1}{x_1}- \frac{1}{x_2}- \frac{1}{x_3}$ will be minimum when $x_1= 3$, $x_2= 4$, and $x_3= 5$. In that case, $1- \frac{1}{x_1}- \frac{1}{x_2}- \frac{1}{x_3}= 1- \frac{1}{3}- \frac{1}{4}- \frac{1}{5}= \frac{60}{60}- \frac{20}{60}- \frac{15}{60}- \frac{12}{60}= \frac{13}{60}$

- Thread starter
- #3

- Nov 26, 2013

- 719

Your answer is not correct. Also please note, that any contribution should be hidden by spoiler tags. Thankyou.

- Apr 22, 2018

- 251

- Thread starter
- #5

- Nov 26, 2013

- 719

You´re right, Olinguito

Thankyou for your participation!