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  $\frac {1}{1/x+1/y+1/z}\leq\frac 19$ .
 A: By AM-HM, we have $\frac 13=\frac {x+y+z}3\geq\frac 3{1/x+1/y+1/z}$, then $\frac 1x+\frac 1y+\frac 1z\geq9$. Equality holds only if $x=y=z=\frac 13$.
A: I have an interesting graphical solution to this. A similar approach is used here
The graph of $y=\frac1x$ looks like this:

I am changing the scale on x-axis for a better look.
Now consider the points $(x,\frac1x)$, $(y,\frac1y)$, $(z,\frac1z)$. The centroid of the triangle formed lies on the line $x=\frac13$ because $x+y+z=1$. Due to the concavity of the graph, the triangle formed will always be above the graph, and hence, thee centroid will also be above it.

So the least value of the ordinate of the centroid would be when all three points coincide and the centroid lands on the graph. In all other cases it will be above the graph.
Thus you obtain the least value when all the numbers are equal
A: 
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}.
