Superadditive Function Analysis for x > 1

A function $f(.)$ is defined as superadditive if it satisfies the inequality:

$f(x_1+x_2) \ge f(x_1) + f(x_2)$

We need to determine which of the given options is superadditive for $x$ > 1. Let's analyze each option:

Option 1: Analysis of $f(x) = e^x$

For the function $f(x) = e^x$, we check the superadditive property for $x_1, x_2$ > 1. We need to verify if:

$e^{x_1+x_2} \ge e^{x_1} + e^{x_2}$

Divide both sides by $e^{x_1+x_2}$ (which is always positive):

$1 \ge \frac{e^{x_1}}{e^{x_1+x_2}} + \frac{e^{x_2}}{e^{x_1+x_2}}$

$1 \ge e^{-x_2} + e^{-x_1}$

Since $x_1$ > 1 and $x_2$ > 1, we know that $0 < e^{-x_1} < e^{-1}$ and $0 < e^{-x_2} < e^{-1}$.

Therefore, the sum $e^{-x_1} + e^{-x_2}$ must be less than $e^{-1} + e^{-1} = 2e^{-1}$.

The value of $2e^{-1}$ is approximately $0.7357$.

Since $0.7357 < 1$, the inequality $1 \ge e^{-x_2} + e^{-x_1}$ is always true for $x_1, x_2$ > 1.

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

Option 2: Analysis of $f(x) = \sqrt{x}$

Consider $f(x) = \sqrt{x}$ for $x$ > 1. Let's test with specific values, for example, $x_1 = 2$ and $x_2 = 2$ (both are > 1).

Left side: $f(x_1+x_2) = f(2+2) = f(4) = \sqrt{4} = 2$.

Right side: $f(x_1) + f(x_2) = f(2) + f(2) = \sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

Since $2$ is not greater than or equal to $2\sqrt{2}$ (approximately $2 < 2.828$), the inequality $f(x_1+x_2) \ge f(x_1) + f(x_2)$ fails.

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

Option 3: Analysis of $f(x) = 1/x$

Consider $f(x) = 1/x$ for $x$ > 1. Let $x_1 = 2$ and $x_2 = 2$.

Left side: $f(x_1+x_2) = f(2+2) = f(4) = 1/4$.

Right side: $f(x_1) + f(x_2) = f(2) + f(2) = 1/2 + 1/2 = 1$.

Since $1/4$ is not greater than or equal to $1$, the inequality $f(x_1+x_2) \ge f(x_1) + f(x_2)$ fails.

Therefore, $f(x) = 1/x$ is not superadditive.

Option 4: Analysis of $f(x) = e^{-x}$

Consider $f(x) = e^{-x}$ for $x$ > 1. Let $x_1 = 2$ and $x_2 = 2$.

Left side: $f(x_1+x_2) = f(2+2) = f(4) = e^{-4}$.

Right side: $f(x_1) + f(x_2) = f(2) + f(2) = e^{-2} + e^{-2} = 2e^{-2}$.

Since $e^{-4}$ (approximately 0.0183) is clearly less than $2e^{-2}$ (approximately 0.2707), the inequality $f(x_1+x_2) \ge f(x_1) + f(x_2)$ fails.

Therefore, $f(x) = e^{-x}$ is not superadditive.

Conclusion

The only function that satisfies the superadditive property $f(x_1+x_2) \ge f(x_1) + f(x_2)$ for $x$ > 1 among the given options is $f(x) = e^x$.