Find the smallest possible value 1-1/(n_1)-1/(n_2)-1/(n_3)

[lfdahl]

Find, with proof, the smallest possible value of $1-\frac{1}{n_1}-\frac{1}{n_2}-\frac{1}{n_3}$

where $n_1,n_2$ and $n_3$ are different positive integers, that satisfy:$\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3} < 1.$

 

[HOI]

First, $1- \frac{1}{x_1}- \frac{1}{x_2}- \frac{1}{x_3}= 1- \left(\frac{1}{x_1}
+ \frac{1}{x_2}+ \frac{1}{x_3}\right)$ 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, $1- \frac{1}{x_1}- \frac{1}{x_2}- \frac{1}{x_3}= 1- \frac{1}{3}- \frac{1}{4}- \frac{1}{5}= \frac{60}{60}- \frac{20}{60}- \frac{15}{60}- \frac{12}{60}= \frac{13}{60}$

 

[lfdahl]

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

 

[Olinguito]

Spoiler

The problem involves maximizing $\dfrac1{n_1}+\dfrac1{n_2}+\dfrac1{n_3}$. We can thus let $n_1=2$ and $n_2=3$. Then $\dfrac12+\dfrac13+\dfrac1{n_3}<1$ $\implies$ $\dfrac1{n_3}<\dfrac16$ $\implies$ $n_3=7$. So the minimum value of $1-\left(\dfrac1{n_1}+\dfrac1{n_2}+\dfrac1{n_3}\right)$ is $1-\left(\dfrac12+\dfrac13+\dfrac17\right)=\dfrac1{42}$.

 

[lfdahl]

Spoiler

The problem involves maximizing $\dfrac1{n_1}+\dfrac1{n_2}+\dfrac1{n_3}$. We can thus let $n_1=2$ and $n_2=3$. Then $\dfrac12+\dfrac13+\dfrac1{n_3}<1$ $\implies$ $\dfrac1{n_3}<\dfrac16$ $\implies$ $n_3=7$. So the minimum value of $1-\left(\dfrac1{n_1}+\dfrac1{n_2}+\dfrac1{n_3}\right)$ is $1-\left(\dfrac12+\dfrac13+\dfrac17\right)=\dfrac1{42}$.

You´re right, Olinguito (Yes)

Thankyou for your participation!