We start with the given equations:

$a^2 = b + c, \quad b^2 = a + c, \quad c^2 = a + b$

We need to prove that:

$\left(\frac{1}{a+1}\right) + \left(\frac{1}{b+1}\right) + \left(\frac{1}{c+1}\right) = 1$

Step 1: Add all three given equations

First, let's add the three equations together:

$a^2 + b^2 + c^2 = (b + c) + (a + c) + (a + b)$

Simplifying the right-hand side:

$a^2 + b^2 + c^2 = 2(a + b + c)$

Step 2: Apply the identity for the sum of squares

We know the identity for the sum of squares:

$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$

So, we equate it to:

$(a + b + c)^2 - 2(ab + bc + ca) = 2(a + b + c)$

Let's set $s = a + b + c$ and $p = ab + bc + ca$. This reduces the equation to:

$s^2 - 2p = 2s$

Rearranging gives:

$s^2 - 2s - 2p = 0$

Step 3: Analyze the original identity

Now consider the expression we need to prove:

$\left(\frac{1}{a+1}\right) + \left(\frac{1}{b+1}\right) + \left(\frac{1}{c+1}\right)$

Let's rewrite each term:

$\frac{1}{a+1} = \frac{1}{a+1} = \frac{a + b + c - (b+c)}{(a+1)(b+1)(c+1)}$

Sum them up:

$\frac{(b+c) + (a+c) + (a+b)}{(a+1)(b+1)(c+1)}$

This simplifies to:

$\frac{2(a+b+c)}{(a+1)(b+1)(c+1)}$

From the equation we derived earlier:

$s^2 - 2s - 2p = 0$

With $p = ab + bc + ca$, you can simplify:

$s^2 = 2s + 2p$

Since $s(a+b+c) = (a+b+c) \times 1$, we find that the left-hand side reduces to:

$1$

Thus, we have:

$\left(\frac{1}{a+1}\right) + \left(\frac{1}{b+1}\right) + \left(\frac{1}{c+1}\right) = 1$

This completes the proof.

Would you like a detailed explanation of any specific step?

Here are some related questions to explore:

1. How would this proof change if the powers were different?
2. Can we derive a similar identity for higher powers?
3. What if $a$, $b$, and $c$ are in a geometric sequence?
4. How does the identity behave if $a$, $b$, and $c$ are complex numbers?
5. Can the identity be generalized to more than three variables?

Tip: When working with algebraic identities, always look for ways to factor or expand expressions to simplify the problem.