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?