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The inequality to be proven is:

\[
(x + y + z) \cdot 2 \left( \frac{1}{2} + \frac{1}{x^2 + y^2} + \frac{1}{y^2 + z^2} + \frac{1}{z^2 + x^2} \right) + \frac{36}{x + y + z} \geq \frac{63}{4}
\]

### Given:
- \(x, y, z\) are non-negative real numbers.
- No two of \(x, y, z\) are equal to zero.

#### Approach:
1. **Simplify the Expression**: First, you can consider simplifying the expression by factoring out common terms and finding symmetry within the variables.
   
2. **Use Symmetry and Inequalities**: Since the expression involves sums of squares and reciprocal terms, you could apply classical inequalities like the **AM-GM inequality**, **Cauchy-Schwarz inequality**, or **Titu's lemma** to get bounds on the terms.

3. **Estimate the Bound**: You want to show that the left-hand side is always greater than or equal to \(\frac{63}{4}\). Focus on the individual terms that might lead to a contradiction if the inequality does not hold, and try using substitutions for easier terms such as letting \(x = y = z\) as a special case to test.

### Suggestions for Next Steps:
- Do you want a detailed step-by-step solution using one of the classical inequalities?
- Would you like assistance with a specific part of the inequality manipulation?

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### Related Questions:
1. How can the AM-GM inequality help simplify expressions with sums of squares?
2. What are the applications of the Cauchy-Schwarz inequality in inequalities involving multiple terms?
3. Can we test this inequality by setting specific values for \(x, y, z\) and see if it holds?
4.

Bài 4. Cho các số thực không âm x, y, z sao cho không có hai số nào bằng 0, chứng minh: (x + y + z) 2 (1/2 + 1/(x^2 + y^2) + 1/(y^2 + z^2) + 1/(z^2 + x^2)) + 36/(x + y + z) >= 63/4

Learn how to prove the inequality involving non-negative real numbers x, y, z where no two numbers are equal to zero. This problem requires the application of advanced inequalities such as the AM-GM inequality and Cauchy-Schwarz inequality.

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