Let $x,$ $y,$ and $z$ be positive real numbers such that $x + y + z = 6.$  Find the minimum value of
\[\frac{x^2 + y^2}{x + y} + \frac{x^2 + z^2}{x + z} + \frac{y^2 + z^2}{y + z}.\]
By QM-AM,
\[\sqrt{\frac{x^2 + y^2}{2}} \ge \frac{x + y}{2}.\]Then
\[\frac{x^2 + y^2}{2} \ge \left( \frac{x +  y}{2} \right)^2,\]which we can re-arrange as
\[\frac{x^2 + y^2}{x + y} \ge \frac{x + y}{2}.\]Similarly,
\begin{align*}
\frac{x^2 + y^2}{x + y} &\ge \frac{x + y}{2}, \\
\frac{y^2 + z^2}{y + z} &\ge \frac{y + z}{2}.
\end{align*}Therefore,
\[\frac{x^2 + y^2}{x + y} + \frac{x^2 + z^2}{x + z} + \frac{y^2 + z^2}{y + z} \ge \frac{x + y}{2} + \frac{x + z}{2} + \frac{y + z}{2} = x + y + z = 6.\]Equality occurs when $x = y = z = 2,$ so the minimum value is $\boxed{6}.$
6
Level 3
Intermediate Algebra
Problem: Let $x,$ $y,$ and $z$ be positive real numbers such that $x + y + z = 6.$  Find the minimum value of
\[\frac{x^2 + y^2}{x + y} + \frac{x^2 + z^2}{x + z} + \frac{y^2 + z^2}{y + z}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the given expression, we first rewrite it in a more manageable form. The expression consists of three terms, each involving the squares of the variables divided by their sums. We can apply the Cauchy-Schwarz inequality to each term.

Using Cauchy-Schwarz, we have:

\[
\frac{x^2 + y^2}{x + y} \geq \frac{(x + y)^2}{x + y} = x + y,
\]
\[
\frac{x^2 + z^2}{x + z} \geq \frac{(x + z)^2}{x + z} = x + z,
\]
\[
\frac{y^2 + z^2}{y + z} \geq \frac{(y + z)^2}{y + z} = y + z.
\]

Adding these inequalities gives:

\[
\frac{x^2 + y^2}{x + y} + \frac{x^2 + z^2}{x + z} + \frac{y^2 + z^2}{y + z} \geq (x + y) + (x + z) + (y + z) = 2(x + y + z).
\]

Since we know that \(x + y + z = 6\), we can substitute this into our inequality:

\[
\frac{x^2 + y^2}{x + y} + \frac{x^2 + z^2}{x + z} + \frac{y^2 + z^2}{y + z} \geq 2 \cdot 6 = 12.
\]

However, we need to find the minimum value of the original expression, not just an upper bound. To achieve this, we can explore specific values of \(x\), \(y\), and \(z\) that satisfy the constraint \(x + y + z = 6\).

Let us consider the case where \(x = y = z = 2\). Substituting these values into the expression yields:

\[
\frac{2^2 + 2^2}{2 + 2} + \frac{2^2 + 2^2}{2 + 2} + \frac{2^2 + 2^2}{2 + 2} = \frac{4 + 4}{4} + \frac{4 + 4}{4} + \frac{4 + 4}{4} = 2 + 2 + 2 = 6.
\]

This value of 6 is achievable and satisfies the conditions of the problem. To confirm that this is indeed the minimum, we can analyze the behavior of the expression as \(x\), \(y\), and \(z\) deviate from 2. 

If we let one variable increase while the others decrease, the terms in the expression will increase due to the quadratic nature of the numerators, leading to values greater than 6. Thus, the minimum value of the expression occurs when \(x = y = z = 2\).

Therefore, the final answer is 6.<|END_OF_LATENT|> The final answer is $6$.