Let $a, b, c$ be real numbers, $a \neq 0$. If the equation $2ax^2 + bx + c = 0$ has real root on the interval $[-1, 1]$.
Prove that 
$$\min \{c, a + c + 1\} \leq \max \{|b - a + 1|, |b + a - 1|\},$$
and determine the necessary and sufficient conditions of $a, b, c$ for the equality case to be achieved.

1. Let \( t \) be a real root of the equation \( 2ax^2 + bx + c = 0 \) in the interval \([-1, 1]\). This implies \( |t| \leq 1 \).

2. Rewriting the equation \( 2ax^2 + bx + c = 0 \) at \( x = t \), we get:
   \[
   2at^2 + bt + c = 0
   \]

3. We can rearrange this equation as:
   \[
   -(a-1)t^2 - bt = (a+1)t^2 + c
   \]

4. Taking the absolute value of both sides, we have:
   \[
   |-(a-1)t^2 - bt| = |(a+1)t^2 + c|
   \]

5. Notice that \( 0 \leq t^2 \leq 1 \), so:
   \[
   |(a+1)t^2 + c| \geq \min \{c, a+1+c\}
   \]

6. Also, since \( -1 \leq t \leq 1 \), we have:
   \[
   |-(a-1)t - b| \cdot |t| \leq |-(a-1)t - b| \leq \max \{|b + a - 1|, |b - a + 1|\}
   \]

7. Combining these inequalities, we get:
   \[
   \min \{c, a + c + 1\} \leq \max \{|b - a + 1|, |b + a - 1|\}
   \]

8. For the equality case, we need:
   \[
   |(a+1)t^2 + c| = \min \{c, a + c + 1\}
   \]
   and
   \[
   |-(a-1)t - b| = \max \{|b - a + 1|, |b + a - 1|\}
   \]

   This implies that \( t \) must be such that both sides of the equation achieve their respective minimum and maximum values.

The final answer is \( \boxed{ \min \{c, a + c + 1\} \leq \max \{|b - a + 1|, |b + a - 1|\} } \)