We aim to prove the inequality:

$\frac{e^{\frac{\ln(x)}{n}} - 1}{\frac{\ln(x)}{n}} \leq e^{\frac{\ln(x)}{n}}$

Step 1: Define the variable

Let $z = \frac{\ln(x)}{n}$. Substituting $z$ into the inequality, it becomes:

$\frac{e^z - 1}{z} \leq e^z$

where $z \neq 0$ (since division by zero is undefined). Now, we analyze this inequality.

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Step 2: Rewrite the inequality

Multiply through by $z$, assuming $z > 0$ (if $z < 0$, we'll analyze separately):

$e^z - 1 \leq z e^z.$

Rearranging terms:

$e^z - z e^z - 1 \leq 0.$

Factoring $e^z$ out from the terms where possible:

$e^z (1 - z) - 1 \leq 0.$

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Step 3: Analyze $f(z) = e^z (1 - z) - 1$

Let $f(z) = e^z (1 - z) - 1$. We will determine whether $f(z) \leq 0$ for all $z$.

1. Compute $f(0)$:

   $f(0) = e^0 (1 - 0) - 1 = 0.$

2. Compute the derivative $f'(z)$:

   Using the product rule:

   $f'(z) = \frac{d}{dz} \left(e^z (1 - z)\right) = e^z (1 - z) + e^z (-1) = e^z (1 - z - 1).$

   Simplifying:

   $f'(z) = -z e^z.$

3. Analyze $f'(z)$:

   • For $z > 0$, $f'(z) = -z e^z < 0$ (decreasing function).
   • For $z < 0$, $f'(z) = -z e^z > 0$ (increasing function).

   Thus, $f(z)$ reaches a maximum at $z = 0$, where $f(0) = 0$.

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Step 4: Conclude

Since $f(z)$ is decreasing for $z > 0$ and increasing for $z < 0$, with $f(0) = 0$, we have $f(z) \leq 0$ for all $z$. This implies:

$e^z (1 - z) - 1 \leq 0,$

or equivalently:

$e^z - 1 \leq z e^z.$

Dividing through by $z > 0$ (or equivalently taking the limit as $z \to 0$), we confirm that:

$\frac{e^z - 1}{z} \leq e^z.$

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Step 5: Original inequality

Returning to the original variables, $z = \frac{\ln(x)}{n}$, we conclude:

$\frac{e^{\frac{\ln(x)}{n}} - 1}{\frac{\ln(x)}{n}} \leq e^{\frac{\ln(x)}{n}}.$

Summary

The inequality holds for all $x > 0$ and $n > 0$. Let me know if you want further clarification!

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Relative Questions:

1. Can you graph $\frac{e^z - 1}{z}$ and $e^z$ to visualize their relationship?
2. How does this proof rely on properties of the exponential function?
3. What happens when $z \to 0$? Does the inequality still hold?
4. How can this inequality be applied in real-world scenarios?
5. Is there a similar inequality involving other transcendental functions?

Tip: For inequalities involving exponential or logarithmic functions, using derivatives to analyze monotonicity is often a powerful tool!