$x,y$ 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.
 A: 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 $3\cdot 135 = 405$.
A: Just do it.  If $z\le  {3*45} $ than $\frac 1x+\frac 1y+\frac 1z >3\frac 1 z\ge \frac 1 {45}$ any such $z-2=x <z-1=y <z\le 135$ will do and the largest $x+y+z $ will be when $z=135$.
Alternatively if $x\ge 135$ then $\frac 1x+\frac 1y+\frac 1z <3\frac 1 x\le \frac 1 {45}$ so none with $ z=x+2>y=x+1>x\ge 135$ will work.
So it's just a matter of seeing if $x=134,y=135,z=136$ will be in range. Or in other words whether $\frac 1{134}+\frac1 {136} >\frac 2 {135} $ or not.
$ \frac 1{134}+\frac1 {136}=\frac {136+134}{(135-1)(135+1)}=\frac {2*135}{135^2-1}>\frac {2*135}{135^2}=\frac {2}{135} $ so indeed $\frac 1{134}+\frac1 {135}+\frac1 {136}>\frac 1 {45} $.
And that's the largest the terms can be.
So the largest $x+y+z=134+135+136=3*135=405$.
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A take away from this seems to be that $\frac 1 {avg(x_i)}\le avg(\frac 1 {x_i}) $.  Can we prove that?
