If a^2=b+c, b^2=a+c, c^2=a+b, prove {(1/a+1)+ (1/b+1)+(1/c+1)}=1 

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.

If a^2 = b + c, b^2 = a + c, c^2 = a + b, prove {(1/a+1)+ (1/b+1)+(1/c+1)}=1

Explore a detailed proof of the identity involving quadratic equations where a^2 = b + c, b^2 = a + c, and c^2 = a + b. The solution demonstrates key algebraic concepts like the sum of squares and symmetric identities, making it suitable for students in Grades 9-12.

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