If `(x-7) (y-10) (z- 12)= 1000`, then find the minimum value of `(x+ y+ z)`

Answer

To find the minimum value of \( x + y + z \) given the equation \( (x - 7)(y - 10)(z - 12) = 1000 \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Rewrite the equation**:
\[ (x - 7)(y - 10)(z - 12) = 1000 \]
Let \( a = x - 7 \), \( b = y - 10 \), and \( c = z - 12 \). Then we have:
\[ abc = 1000 \]

2. **Apply the AM-GM inequality**:
The AM-GM inequality states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
Substituting \( abc = 1000 \):
\[ \frac{a + b + c}{3} \geq \sqrt[3]{1000} \]

3. **Calculate the cube root**:
\[ \sqrt[3]{1000} = 10 \]
Therefore:
\[ \frac{a + b + c}{3} \geq 10 \]

4. **Multiply both sides by 3**:
\[ a + b + c \geq 30 \]

5. **Substitute back to find \( x + y + z \)**:
Recall that:
\[ a = x - 7, \quad b = y - 10, \quad c = z - 12 \]
Thus:
\[ (x - 7) + (y - 10) + (z - 12) = a + b + c \]
This simplifies to:
\[ x + y + z - 29 \geq 30 \]

6. **Add 29 to both sides**:
\[ x + y + z \geq 30 + 29 \]
\[ x + y + z \geq 59 \]

7. **Conclusion**:
The minimum value of \( x + y + z \) is \( 59 \).