Understanding Superadditive Functions

A function \(f(x)\) is defined as superadditive if for all \(x_1, x_2\) in its domain, the following inequality holds:

\(f\left( {{x_1} + {x_2}} \right) \ge f\left( {{x_1}} \right) + f\left( {{x_2}} \right)\)

We need to determine which of the given functions satisfies this property for values of \(x > 1\). This means we consider \(x_1 > 1\) and \(x_2 > 1\).

Analyzing the Exponential Function \(e^x\)

Let's test if \(f(x) = e^x\) is a superadditive function for \(x > 1\).

We need to check if \(e^{x_1 + x_2} \ge e^{x_1} + e^{x_2}\) for \(x_1, x_2 > 1\).

We know that \(e^{x_1 + x_2} = e^{x_1} \cdot e^{x_2}\).

So the inequality we need to verify becomes \(e^{x_1} \cdot e^{x_2} \ge e^{x_1} + e^{x_2}\).

To prove this, we can rearrange the terms:

• Subtract \(e^{x_1}\) and \(e^{x_2}\) from both sides:
  \(e^{x_1}e^{x_2} - e^{x_1} - e^{x_2} \ge 0\)
• Add 1 to both sides of the inequality:
  \(e^{x_1}e^{x_2} - e^{x_1} - e^{x_2} + 1 \ge 1\)
• Factor the left side of the inequality. This is a common factoring pattern similar to \((a-1)(b-1) = ab - a - b + 1\):
  \((e^{x_1} - 1)(e^{x_2} - 1) \ge 1\)

Now, let's consider the conditions \(x_1 > 1\) and \(x_2 > 1\).

• Since \(x_1 > 1\), the value of \(e^{x_1}\) will be greater than \(e^1\), which is approximately 2.718.
  So, \(e^{x_1} > e \approx 2.718\).
• This implies that \(e^{x_1} - 1 > e - 1 \approx 1.718\).
• Similarly, since \(x_2 > 1\), \(e^{x_2} > e \approx 2.718\).
• This implies that \(e^{x_2} - 1 > e - 1 \approx 1.718\).

Now, let's multiply the lower bounds of \((e^{x_1} - 1)\) and \((e^{x_2} - 1)\):

\((e^{x_1} - 1)(e^{x_2} - 1) > (e - 1)(e - 1)\)

\((e^{x_1} - 1)(e^{x_2} - 1) > (1.718)(1.718) \approx 2.95\)

Since \(2.95 \ge 1\), the inequality \((e^{x_1} - 1)(e^{x_2} - 1) \ge 1\) holds true for all \(x_1, x_2 > 1\).

Thus, \(f(x) = e^x\) is a superadditive function for \(x > 1\).

Evaluating the Square Root Function \(\sqrt{x}\)

Let's check if \(f(x) = \sqrt{x}\) is a superadditive function for \(x > 1\).

We need to check if \(\sqrt{x_1 + x_2} \ge \sqrt{x_1} + \sqrt{x_2}\) for \(x_1, x_2 > 1\).

Let's use a simple example. Let \(x_1 = 4\) and \(x_2 = 4\). These values satisfy \(x > 1\).

• Left Hand Side (LHS): \(\sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.828\)
• Right Hand Side (RHS): \(\sqrt{4} + \sqrt{4} = 2 + 2 = 4\)

Comparing the values, \(2.828 \not\ge 4\). The inequality does not hold.

In fact, for any positive \(x_1, x_2\), it is known that \(\sqrt{x_1 + x_2} < \sqrt{x_1} + \sqrt{x_2}\) (unless one is zero). This means that the square root function is actually a subadditive function.

Therefore, \(f(x) = \sqrt{x}\) is not a superadditive function.

Examining the Reciprocal Function \(1/x\)

Let's check if \(f(x) = 1/x\) is a superadditive function for \(x > 1\).

We need to check if \(\frac{1}{x_1 + x_2} \ge \frac{1}{x_1} + \frac{1}{x_2}\) for \(x_1, x_2 > 1\).

Let's use a simple example. Let \(x_1 = 2\) and \(x_2 = 2\). These values satisfy \(x > 1\).

• Left Hand Side (LHS): \(\frac{1}{2 + 2} = \frac{1}{4}\)
• Right Hand Side (RHS): \(\frac{1}{2} + \frac{1}{2} = 1\)

Comparing the values, \(\frac{1}{4} \not\ge 1\). The inequality does not hold.

To generalize, the right side can be written as \(\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2}\).

So, we are checking if \(\frac{1}{x_1 + x_2} \ge \frac{x_1 + x_2}{x_1 x_2}\).

Since \(x_1, x_2 > 1\), all terms are positive. We can cross-multiply:

\(x_1 x_2 \ge (x_1 + x_2)^2\)

\(x_1 x_2 \ge x_1^2 + 2x_1 x_2 + x_2^2\)

Rearranging the terms:

\(0 \ge x_1^2 + x_1 x_2 + x_2^2\)

Since \(x_1 > 1\) and \(x_2 > 1\), \(x_1^2\), \(x_1 x_2\), and \(x_2^2\) are all positive numbers. Their sum \(x_1^2 + x_1 x_2 + x_2^2\) must be positive. Therefore, the inequality \(0 \ge x_1^2 + x_1 x_2 + x_2^2\) is false.

Therefore, \(f(x) = 1/x\) is not a superadditive function. In fact, it is also a subadditive function for positive \(x\).

Investigating the Negative Exponential Function \(e^{-x}\)

Let's check if \(f(x) = e^{-x}\) is a superadditive function for \(x > 1\).

We need to check if \(e^{-(x_1 + x_2)} \ge e^{-x_1} + e^{-x_2}\) for \(x_1, x_2 > 1\).

This can also be written as \(\frac{1}{e^{x_1 + x_2}} \ge \frac{1}{e^{x_1}} + \frac{1}{e^{x_2}}\).

Let's use a simple example. Let \(x_1 = 2\) and \(x_2 = 2\). These values satisfy \(x > 1\).

• Left Hand Side (LHS): \(e^{-(2+2)} = e^{-4} = \frac{1}{e^4}\)
• Right Hand Side (RHS): \(e^{-2} + e^{-2} = \frac{1}{e^2} + \frac{1}{e^2} = \frac{2}{e^2}\)

The inequality becomes \(\frac{1}{e^4} \ge \frac{2}{e^2}\).

Multiply both sides by \(e^4\) (since \(e^4 > 0\)):

\(1 \ge 2e^2\)

We know that \(e \approx 2.718\), so \(e^2 \approx (2.718)^2 \approx 7.389\).
Therefore, \(2e^2 \approx 2 \times 7.389 = 14.778\).

Clearly, \(1 \not\ge 14.778\). The inequality does not hold.

Therefore, \(f(x) = e^{-x}\) is not a superadditive function. Similar to \(1/x\) and \(\sqrt{x}\), it is a subadditive function for positive \(x\).

Summary of Function Superadditivity for \(x > 1\)

Conclusion on Superadditive Functions

Based on our detailed analysis, only the exponential function \(f(x) = e^x\) satisfies the property of a superadditive function for \(x > 1\).