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The image contains two mathematical statements to prove, where \( a \), \( b \), and \( c \) are real numbers that satisfy the equation:

\[
a^2 + b^2 + c^2 = 2
\]

The two parts to prove are:

1. **Prove that if \( a \leq 0 \), then \( a + b + c \leq abc + 2 \).**

2. **Show that \( a + b + c \leq abc + 2 \) for all real numbers \( a \), \( b \), and \( c \).**

### Let’s break these down:

1. **Proof for \( a \leq 0 \):**
   - Given that \( a^2 + b^2 + c^2 = 2 \), we know that each of \( a \), \( b \), and \( c \) is bounded because their squares sum up to a fixed number.
   - We need to analyze what happens when \( a \leq 0 \) and prove the inequality \( a + b + c \leq abc + 2 \).

2. **General proof:**
   - Here, we need to show that the inequality \( a + b + c \leq abc + 2 \) holds for any real numbers \( a \), \( b \), and \( c \).
   - Again, we will leverage the condition \( a^2 + b^2 + c^2 = 2 \) to derive this result.

Would you like me to start by solving part 1 of this proof?

Soient a, b et c des nombres réels tels que a^2 + b^2 + c^2 = 2. 1. Prouver que si a ≤ 0 alors a + b + c ≤ abc + 2. 2. Montrer que a + b + c ≤ abc + 2 pour tout a, b et c dans R.

Learn how to prove algebraic inequalities involving real numbers a, b, and c with the condition a^2 + b^2 + c^2 = 2. The problem includes proving two inequalities and showcases real number analysis and inequality bounds. Suitable for undergraduate-level math students.

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