Let $x,y,z$ be positive real numbers such that $x+y+z=1$  Prove that always $\left( 1+\frac1x\right)\times\left(1+\frac1y\right)\times\left(1 +\frac1z\right)\ge 64$
When does equality hold?
To prove that for positive real numbers \(x, y, z\) such that \(x + y + z = 1\), the inequality 
\[
\left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{y} \right) \left( 1 + \frac{1}{z} \right) \geq 64
\]
holds, we can use the AM-GM inequality.

1. **Rewrite the expression:**
   \[
   \left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{y} \right) \left( 1 + \frac{1}{z} \right) = \left( \frac{x+1}{x} \right) \left( \frac{y+1}{y} \right) \left( \frac{z+1}{z} \right)
   \]

2. **Simplify the expression:**
   \[
   \left( \frac{x+1}{x} \right) \left( \frac{y+1}{y} \right) \left( \frac{z+1}{z} \right) = \frac{(x+1)(y+1)(z+1)}{xyz}
   \]

3. **Expand the numerator:**
   \[
   (x+1)(y+1)(z+1) = xyz + xy + yz + zx + x + y + z + 1
   \]

4. **Substitute \(x + y + z = 1\):**
   \[
   (x+1)(y+1)(z+1) = xyz + xy + yz + zx + 1 + 1 = xyz + xy + yz + zx + 2
   \]

5. **Substitute back into the fraction:**
   \[
   \frac{(x+1)(y+1)(z+1)}{xyz} = \frac{xyz + xy + yz + zx + 2}{xyz} = 1 + \frac{xy + yz + zx}{xyz} + \frac{2}{xyz}
   \]

6. **Simplify the expression:**
   \[
   1 + \frac{xy + yz + zx}{xyz} + \frac{2}{xyz} = 1 + \frac{xy + yz + zx + 2}{xyz}
   \]

7. **Use AM-GM inequality:**
   By AM-GM inequality, we know that for positive real numbers \(a, b, c\),
   \[
   \frac{a + b + c}{3} \geq \sqrt[3]{abc}
   \]
   Applying this to \(x, y, z\):
   \[
   \frac{x + y + z}{3} \geq \sqrt[3]{xyz}
   \]
   Since \(x + y + z = 1\):
   \[
   \frac{1}{3} \geq \sqrt[3]{xyz}
   \]
   Cubing both sides:
   \[
   \left( \frac{1}{3} \right)^3 \geq xyz \implies \frac{1}{27} \geq xyz
   \]

8. **Substitute \(xyz \leq \frac{1}{27}\) into the inequality:**
   \[
   1 + \frac{xy + yz + zx + 2}{xyz} \geq 1 + \frac{xy + yz + zx + 2}{\frac{1}{27}}
   \]
   \[
   = 1 + 27(xy + yz + zx + 2)
   \]

9. **Use AM-GM on \(xy + yz + zx\):**
   By AM-GM inequality:
   \[
   xy + yz + zx \leq \frac{(x + y + z)^2}{3} = \frac{1^2}{3} = \frac{1}{3}
   \]

10. **Substitute \(xy + yz + zx \leq \frac{1}{3}\):**
    \[
    1 + 27 \left( \frac{1}{3} + 2 \right) = 1 + 27 \left( \frac{1}{3} + \frac{6}{3} \right) = 1 + 27 \left( \frac{7}{3} \right) = 1 + 63 = 64
    \]

Thus, we have shown that:
\[
\left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{y} \right) \left( 1 + \frac{1}{z} \right) \geq 64
\]

Equality holds when \(x = y = z = \frac{1}{3}\).

The final answer is \( \boxed{ 64 } \)