If x,y,z are three +ive real numbers, then minimum value of `(y+z)/x+(z+x)/y+(x+y)/z` is

Options

A. 1
B. 2
C. 3
D. 6

Answer

To find the minimum value of the expression \(\frac{y+z}{x} + \frac{z+x}{y} + \frac{x+y}{z}\) where \(x, y, z\) are positive real numbers, we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-step solution:

1. **Rewrite the expression**:
\[ E = \frac{y+z}{x} + \frac{z+x}{y} + \frac{x+y}{z} \]

2. **Apply AM-GM Inequality**:
We can apply the AM-GM inequality to each of the terms in the expression:
\[ \frac{y+z}{x} = \frac{y}{x} + \frac{z}{x} \]
By AM-GM, we know that:
\[ \frac{y}{x} + \frac{z}{x} \geq 2\sqrt{\frac{yz}{x^2}} \]
Similarly, we can apply AM-GM to the other two terms:
\[ \frac{z+x}{y} \geq 2\sqrt{\frac{zx}{y^2}} \]
\[ \frac{x+y}{z} \geq 2\sqrt{\frac{xy}{z^2}} \]

3. **Combine the inequalities**:
Adding these inequalities together gives:
\[ E \geq 2\left(\sqrt{\frac{yz}{x^2}} + \sqrt{\frac{zx}{y^2}} + \sqrt{\frac{xy}{z^2}}\right) \]

4. **Use AM-GM again**:
Now, we can apply AM-GM to the terms \(\sqrt{\frac{yz}{x^2}}, \sqrt{\frac{zx}{y^2}}, \sqrt{\frac{xy}{z^2}}\):
\[ \sqrt{\frac{yz}{x^2}} + \sqrt{\frac{zx}{y^2}} + \sqrt{\frac{xy}{z^2}} \geq 3\sqrt[3]{\sqrt{\frac{yz}{x^2}} \cdot \sqrt{\frac{zx}{y^2}} \cdot \sqrt{\frac{xy}{z^2}}} \]
Simplifying this gives:
\[ = 3\sqrt[3]{\frac{(xyz)^{1.5}}{(xyz)}} = 3 \]

5. **Final inequality**:
Thus, we have:
\[ E \geq 2 \cdot 3 = 6 \]

6. **Equality condition**:
The equality in AM-GM holds when \(x = y = z\). Therefore, the minimum value of \(E\) occurs when \(x = y = z\).

### Conclusion:
The minimum value of \(\frac{y+z}{x} + \frac{z+x}{y} + \frac{x+y}{z}\) is \(6\).