Let $x,$ $y,$ and $z$ be nonnegative real numbers such that $x + y + z = 5.$  Find the maximum value of
\[\sqrt{2x + 1} + \sqrt{2y + 1} + \sqrt{2z + 1}.\]
By QM-AM,
\[\sqrt{\frac{(2x + 1) + (2y + 1) + (2z + 1)}{3}} \ge \frac{\sqrt{2x + 1} + \sqrt{2y + 1} + \sqrt{2z + 1}}{3}.\]Hence,
\[\sqrt{2x + 1} + \sqrt{2y + 1} + \sqrt{2z + 1} \le \sqrt{3(2x + 2y + 2z + 3)} = \sqrt{39}.\]Equality occurs when $x = y = z = \frac{5}{3},$ so the maximum value is $\boxed{\sqrt{39}}.$
\sqrt{39}
Level 4
Intermediate Algebra
Problem: Let $x,$ $y,$ and $z$ be nonnegative real numbers such that $x + y + z = 5.$  Find the maximum value of
\[\sqrt{2x + 1} + \sqrt{2y + 1} + \sqrt{2z + 1}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the maximum value of the expression, we need to analyze the function \( f(x) = \sqrt{2x + 1} \). This function is concave because its second derivative is negative. By Jensen's inequality, for a concave function, the maximum occurs at the endpoints of the domain.

Given the constraint \( x + y + z = 5 \), we can express \( z \) in terms of \( x \) and \( y \): \( z = 5 - x - y \). The expression we want to maximize can be rewritten as:

\[
f(x) + f(y) + f(5 - x - y) = \sqrt{2x + 1} + \sqrt{2y + 1} + \sqrt{2(5 - x - y) + 1}.
\]

To maximize this, we can use the method of Lagrange multipliers or analyze critical points. However, a simpler approach is to consider the symmetry of the problem. Since \( x, y, z \) are nonnegative and their sum is fixed, we can assume \( x = y = z \) for maximum symmetry.

Setting \( x = y = z = \frac{5}{3} \) gives:

\[
\sqrt{2\left(\frac{5}{3}\right) + 1} = \sqrt{\frac{10}{3} + 1} = \sqrt{\frac{13}{3}}.
\]

Calculating the total:

\[
3 \cdot \sqrt{\frac{13}{3}} = \sqrt{39}.
\]

To confirm this is indeed the maximum, we can check other combinations of \( x, y, z \) that satisfy the constraint. For instance, if we set \( x = 5, y = 0, z = 0 \):

\[
\sqrt{2 \cdot 5 + 1} + \sqrt{1} + \sqrt{1} = \sqrt{11} + 1 + 1 < \sqrt{39}.
\]

Similarly, testing other combinations like \( x = 4, y = 1, z = 0 \) yields values less than \( \sqrt{39} \).

Thus, the maximum value of \( \sqrt{2x + 1} + \sqrt{2y + 1} + \sqrt{2z + 1} \) under the given constraint is indeed \( \sqrt{39} \).<|END_OF_LATENT|> The final answer is $\sqrt{39}$.