Let $x$ and $y$ be positive real numbers such that $x + y = 10.$  Find the minimum value of $\frac{1}{x} + \frac{1}{y}.$
By AM-HM,
\[\frac{x + y}{2} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}}.\]Hence,
\[\frac{1}{x} + \frac{1}{y} \ge \frac{4}{x + y} = \frac{4}{10} = \frac{2}{5}.\]Equality occurs when $x = y = 5,$ so the minimum value is $\boxed{\frac{2}{5}}.$
\frac{2}{5}
Level 2
Intermediate Algebra
Problem: Let $x$ and $y$ be positive real numbers such that $x + y = 10.$  Find the minimum value of $\frac{1}{x} + \frac{1}{y}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the expression $\frac{1}{x} + \frac{1}{y}$ under the constraint $x + y = 10$, we can use the method of Lagrange multipliers or apply the Cauchy-Schwarz inequality. 

First, we rewrite the expression in terms of a single variable. Since $y = 10 - x$, we substitute this into the expression:

$$\frac{1}{x} + \frac{1}{y} = \frac{1}{x} + \frac{1}{10 - x}.$$

Next, we need to find the critical points of this function. We differentiate the expression with respect to $x$:

$$f(x) = \frac{1}{x} + \frac{1}{10 - x}.$$

The derivative is:

$$f'(x) = -\frac{1}{x^2} + \frac{1}{(10 - x)^2}.$$

Setting the derivative equal to zero to find critical points:

$$-\frac{1}{x^2} + \frac{1}{(10 - x)^2} = 0.$$

This implies:

$$\frac{1}{x^2} = \frac{1}{(10 - x)^2}.$$

Cross-multiplying gives:

$$(10 - x)^2 = x^2.$$

Expanding both sides results in:

$$100 - 20x + x^2 = x^2.$$

Simplifying leads to:

$$100 - 20x = 0,$$

which gives:

$$x = 5.$$

Since $x + y = 10$, we also have $y = 5$. 

Now, substituting $x = 5$ and $y = 5$ back into the original expression:

$$\frac{1}{5} + \frac{1}{5} = \frac{2}{5}.$$

To confirm that this is indeed a minimum, we can check the second derivative:

$$f''(x) = \frac{2}{x^3} + \frac{2}{(10 - x)^3}.$$

Since both terms are positive for $x > 0$ and $x < 10$, $f''(x) > 0$ indicates that $f(x)$ is concave up at $x = 5$, confirming a local minimum.

Thus, the minimum value of $\frac{1}{x} + \frac{1}{y}$, given the constraint $x + y = 10$, is $\frac{2}{5}$.<|END_OF_LATENT|> The final answer is $\frac{2}{5}$.