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

Question

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?

Answer 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 $x=y=z=\frac 13$.

Answer 2

I think you're asking why plugging $x+y=z=1$ restricting the equation to that sphere leads to the minimum on the sphere being the global minimum, so you're asking why the minimum couldn't exist when $x+y+z\ne1$.

Assume it exists when $x+y+z=t<1$ then

$$\frac1x+\frac1y+\frac1z\ge \frac9t > 9 $$

Can you explain why the minimum can't be attained when $x+y+z=t>1$?