let $\frac1x +\frac1y+\frac1z < 1$ such that $z ,y , x \in \mathbb{N}$ and $z\neq y \neq x$ then find minimum of $A=1-(\frac1x +\frac1y+\frac1z)$
My Work :
$$\frac1x +\frac1y+\frac1z < 1 \\\frac{yz+xz+xy}{xyz} < 1 \\ yz+xz+xy < xyz $$
I can't . please help me!
On
Consider the set of sums $\frac 1x + \frac 1y + \frac 1z$ for $x,y,z \in \mathbb N$. Clearly, if $x,y,z$ increase, then the sum decreases. Furthermore, the sum is symmetric in $x,y,z$. WLOG assume $x < y < z$. We will go case by case.
(When I refer to sum, or sum so far, in the below paragraphs, I refer to $\frac 1{x} + \frac 1y + \frac 1z$ and $\frac 1x + \frac 1y$ respectively).
CASE 1 : $x = 2$.
Start with $y=3$, then the sum is so far $\frac 56$. So adding $\frac 17$ would make it less than $1$, by precisely $\frac 1{42}$, and for greater $z$ this difference will only get bigger.
Now take $y=4$, the sum is so far $\frac 34$, and since $z \geq 5$, we see that the final sum be less than $1$ by at least $\frac 1{20}$.
Taking $y > 4$ would mean that the final sum is at least smaller than $\frac 12 + \frac 15 + \frac 16$, which is not even $\frac 9{10}$. So we can finish $x=2$ here.
CASE $2$ : $x = 3$.
If $x=3$, then you can check that if $y > 4$ we can obtain a trivial bound for the sum of reciprocals like last time.
If $y = 4$, then the sum so far is $\frac 7{12}$, so $z = 5$ works and gives a final difference from $1$ of $\frac {13}{60}$ which is very large.
CASE $3$ : $x > 3$.
This is obsolete, because then the final sum must be less than $\frac 14 + \frac 15 + \frac 16 = \frac {37}{60}$.
Hence, this case by case analysis gives the minimum to be $\frac 1{42}$.
Hint.
To make $A$ as small as possible you need to make $\frac1x+\frac1y+\frac1z$ as large as possible, but less than $1$. Given all your constraints on $x$, $y$, $z$, and the fact that $\frac12+\frac13+\frac16=1$ is too large, there are not so many options: the answer is either $2$, $3$, $7$, or $2$, $4$, $6$, or numbers all lower than $6$.