<s>[INST] If $x,y, z>0$ and $x+y+z=1$ then prove that $\frac 1x+\frac 1y+\frac 1z≥9$? Can some please help me out with this question?
I tried it in this way
Note that  $a,b>0$  then  $a≥b\iff\frac 1a≤\frac1b$ .
So it is sufficient to prove  $\frac {1}{1/x+1/y+1/z}\leq\frac 19$ .
 [/INST] 
If $x,y,z>0$ and $x+y+z=1$, then prove that
\begin{align}\frac1x+\frac1y+\frac1z\ge9 \tag{1}\label{1}. \end{align}

Alternatively, using Ravi substitution,
where the triplet $a, b, c$ represents the sides of a valid triangle
with semiperimeter $\rho$, inradius $r$ and circumradius $R$,
\begin{align}
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
\tag{2}\label{2}
,
\end{align}
\eqref{1} becomes
\begin{align}
\frac{3\rho^2-2(a+c+b)\,\rho+(ab+bc+ac)}
{\rho^3-(a+b+c)\rho^2+(ab+bc+ac)\,\rho-abc}
&\ge9
\tag{3}\label{3}
,\\
\frac{3\rho^2-2\,(2\rho)\,\rho+(\rho^2+r^2+4rR)}
{\rho^3-(2\,\rho)\rho^2+(\rho^2+r^2+4rR)\,\rho-4\rho r R}
-9
&\ge0
\tag{4}\label{4}
,
\end{align}
\begin{align} 
\frac{4R+r-9\rho r}{\rho r}
&\ge0
\tag{5}\label{5}
.
\end{align}
Since $\rho=x+y+z=1$,
\eqref{5} simplifies to
\begin{align} 
R-2r&\ge0
\tag{6}\label{6}
,
\end{align}
which always holds for a valid triangle.
The equality corresponds only to the equilateral triangle,
$a=b=c$, that is, $x=y=z=\tfrac13$
in the original \eqref{1}.
</s>