# Finding Maximum value of consecutive positive integers with constraints

Let, $x,y,z$ be consecutive positive integers such that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} > \frac{1}{10}$ I need to find the maximum value of $x+y+z$ ?

My attempt:

I tried guessing first, To maximise $x+y+z$ it is necessary to have high value for $x$, But not too high which will contradict the constrain of the problem. As we can roughly split the $0.1$ into three parts so I started with numbers closer to $30$ with trial an error I got that for $x=29$ we get the optimum and final result would be 90 (=max. value of $x+y+z$).

Is there any way to arrive at this conclusion with using inequalities? Any hints will be appreciated...

• Try setting $y=x+1$ and $z=x+2$ Apr 5, 2017 at 19:17

Let $n = 30$, so that we don't get distracted. We're looking for the largest $x \in \mathbb{N}$ such that $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2} > \frac{3}{n}$. Your (right) guess is to check $x$ close to $n$.
Indeed, for $x = n$ we have $$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2} < \frac{1}{n}+\frac{1}{n}+\frac{1}{n} = \frac{3}{n},$$ while for $x =n-1$ $$\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1} > \frac{3}{n}.$$ The above inequality comes from $\frac{1}{n-1}+\frac{1}{n+1} = \frac{2n}{n^2-1} > \frac{2}{n}$.