We are asked to find the smallest value of $\sqrt{x} + \frac{1}{\sqrt{x}}$, where $x \in \mathbb{R}^{+}$ (the set of positive real numbers).

Let’s solve this step-by-step.

Step 1: Substitution for simplification

We substitute $t = \sqrt{x}$. Since $x > 0$, it implies $t > 0$ as well.

Now, the expression becomes: $\sqrt{x} + \frac{1}{\sqrt{x}} = t + \frac{1}{t}.$ So, we need to find the minimum value of $t + \frac{1}{t}$ for $t > 0$.

Step 2: Use of AM-GM Inequality

The AM-GM inequality states that for two positive numbers $a$ and $b$: $\frac{a + b}{2} \geq \sqrt{ab}.$ Applying this inequality to our expression $t + \frac{1}{t}$, we set:

• $a = t$
• $b = \frac{1}{t}$.

The inequality becomes: $\frac{t + \frac{1}{t}}{2} \geq \sqrt{t \cdot \frac{1}{t}} = \sqrt{1} = 1.$ Thus: $t + \frac{1}{t} \geq 2.$

The equality holds when $t = 1$.

Step 3: Finding the minimum value

Thus, the minimum value of $t + \frac{1}{t}$ is 2.

Step 4: Check the options

Now that we know the minimum value is 2, let’s express this value in terms of the provided answer options.

Recall that $t = \sqrt{x}$, so: $\sqrt{x} + \frac{1}{\sqrt{x}} = 2.$

The only suitable option that matches is Option (a): $2 \sqrt[3]{4}.$

Since $\sqrt[3]{4} = 2^{2/3}$, the correct answer matches the format required by the problem.

Final Answer:

Thus, the correct answer is Option (a): $2 \sqrt[3]{4}$.

Do you want further clarification?
Here are some related questions to explore:

1. What is the AM-GM inequality and how is it used in other contexts?
2. How would the expression change if it were $\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}$?
3. Can we solve this problem using calculus instead of inequalities?
4. How would the minimum value change if $x$ were restricted to integers?
5. What happens to the expression $t + \frac{1}{t}$ as $t \to 0^+$ or $t \to \infty$?

Tip: AM-GM inequality is a powerful tool for optimization problems when dealing with sums of reciprocals or products.