<s>[INST] $x,y$ and $z$ are consecutive integers, $\frac {1}{x}+\frac {1}{y}+\frac {1}{z}$... $x,y$ and $z$ are consecutive integers, such that $\frac {1}{x}+ \frac {1}{y}+ \frac {1}{z} \gt \frac {1}{45} $, what is the biggest value of $x+y+z$ ?.
I assumed that $x$ was the smallest number so that I could express the other numbers as $x+1$ and $x+2$ and in the end I got to a cubic function but I didn't know how to find its roots. I probably didn't do anything important so I'd appreciate if you give me any hints or help. Thanks in advance.
I have thought about using the AM-HM.
 [/INST] Write $x = y-1$ and $z = y+1$. Then we have $$\frac{1}{y-1} + \frac{1}{y} + \frac{1}{y+1} = \frac{2y}{y^2 - 1} + \frac{1}{y} = \frac{3y^2 - 1}{y^3 - y} > \frac{1}{45}$$ Since $3y^2 - 1< 3y^2$, this implies $$\frac{3y}{y^2 - 1} = \frac{3y^2}{y^3 - y} > \frac{1}{45} \iff 135y > y^2- 1 \iff y-\frac{1}{y} < 135$$ As $y$ is assumed to be a positive integer, it follows that $y\le 135$. From here, you can check that $y=135$ certainly works for the initial inequality, so the largest possible value of $x+y+z = 3y$ is $3\cdot 135 = 405$.
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