`2/(b+c)+2/(c+a)+2/(a+b)gt9/(a+b+c)`

Answer

To solve the inequality \( \frac{2}{b+c} + \frac{2}{c+a} + \frac{2}{a+b} > \frac{9}{a+b+c} \), we can use the concept of the Arithmetic Mean-Harmonic Mean (AM-HM) inequality.

### Step-by-step solution:

1. **Identify the terms**: We have three fractions on the left side:
\[ \frac{2}{b+c}, \quad \frac{2}{c+a}, \quad \frac{2}{a+b} \]

2. **Apply the AM-HM inequality**: According to the AM-HM inequality, for any positive numbers \( x_1, x_2, x_3 \):
\[ \frac{x_1 + x_2 + x_3}{3} \geq \frac{3}{\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3}} \]
Here, we can set:
\[ x_1 = b+c, \quad x_2 = c+a, \quad x_3 = a+b \]

3. **Calculate the left-hand side**:
\[ \frac{(b+c) + (c+a) + (a+b)}{3} = \frac{2(a+b+c)}{3} \]

4. **Calculate the right-hand side using HM**:
\[ \frac{3}{\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}} \Rightarrow \text{Let } S = \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \]

5. **Combine the results**:
From the AM-HM inequality, we have:
\[ \frac{2(a+b+c)}{3} \geq \frac{3}{S} \]

6. **Rearranging gives**:
\[ S \geq \frac{9}{2(a+b+c)} \]

7. **Substituting back into the inequality**:
We need to show:
\[ \frac{2}{b+c} + \frac{2}{c+a} + \frac{2}{a+b} > \frac{9}{a+b+c} \]
This can be rewritten using \( S \):
\[ 2S > \frac{9}{a+b+c} \]

8. **Final step**:
Since we have \( S \geq \frac{9}{2(a+b+c)} \), multiplying both sides by 2 gives:
\[ 2S \geq \frac{9}{a+b+c} \]
Thus, we conclude:
\[ \frac{2}{b+c} + \frac{2}{c+a} + \frac{2}{a+b} > \frac{9}{a+b+c} \]
This proves the original inequality.