Minimum of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ for $x,y,z\in{\mathbb{R^+}}$

I may be overthinking but my question is this: I've proved the inequality $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)≥9$ and then I've been asked if the minimum of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ equal to $9$ given that $x+y+z=1$. Clearly if I just plug $x+y+z=1$ into the given inequality then yes the minimum will be $9$ but I have a feeling there is a restriction somewhere that stops this but I can't identify where because I don't know how I could write $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ with an $x+y+z$ in it so I'm not sure how the two expressions would relate. So am I overthinking and is the minimum $9$ or have I missed something?

Solution 1:

You just proved that 9 is a lower bound ($\frac 1x+ \frac 1y+ \frac 1z \geq 9$). You then have to find a triple $(x,y,z)$ such that $x+y+z=1$ and $\frac 1x+ \frac 1y+ \frac 1z =9$. Such a triple is $x=y=z=\frac 13$.