Minimum Value of Expression

Let the $A=\frac{x+y+z}{3}$ be the arithmetic mean of $x,y,z$ and let $H=\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$ be the harmonic mean. Then $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) = 9\frac{A}{H}$ For positive real numbers, $A\geq H$ with equality holding precisely when $x=y=z$ ( see here for a proof ). Since $\frac{A}{H}$ will be minimized when equality holds, $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) = 3x\left(\frac{3}{x}\right) = \boxed{9}$

Using Hölder's inequality as follows:

$\begin{aligned} (x+y+z)^\frac 12\left(\frac 1x+\frac 1y + \frac 1z\right)^\frac 12 & \ge 1 + 1 + 1 & \small \color{#3D99F6} \text{Squaring both sides} \\ (x+y+z)\left(\frac 1x+\frac 1y + \frac 1z\right) & \ge (1 + 1 + 1)^2 = \boxed 9 \end{aligned}$

By AM-GM inequality, we have $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \ge 3(xyz)^{\frac{1}{3}} \times 3(\frac{1}{xyz})^{\frac{1}{3}}=9$ The equality holds iff $x=y=z$

To get the smallest numbers possible on both the LHS and RHS, we need to make x = y = z. The miraculous thing about doing this is that no matter what value you substitute for x, y or z, you always get 9. Here is the proof:

(x + y + z)(1/x +1/y + 1/z)

1 + x/y + x/z + y/x + 1 + y/z + z/x + z/y + 1

Since x = y = z, all terms that contain these variables simplify to 1.

Therefore, we get 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9

Cauchy Schwarz inequality