1. Let a1, a2, ..., an be positive real numbers. Prove by mathematical indution that the arithmetic
mean of these numbers is greater or equals to the geometric mean.

We are tasked with proving the **Arithmetic Mean - Geometric Mean Inequality (AM-GM Inequality)** using mathematical induction.

The inequality states that for any set of positive real numbers \(a_1, a_2, \dots, a_n\), the arithmetic mean is greater than or equal to the geometric mean:

\[
\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot \dots \cdot a_n}
\]

with equality holding if and only if \(a_1 = a_2 = \dots = a_n\).

### Base Case: \(n = 1\)

For \(n = 1\), the arithmetic mean and geometric mean are both just the number itself:
\[
\frac{a_1}{1} = a_1 \quad \text{and} \quad \sqrt[1]{a_1} = a_1
\]
Clearly, \(\frac{a_1}{1} = a_1 \geq a_1\), so the inequality holds for \(n = 1\).

### Inductive Step

Now, assume the inequality holds for some \(n = k\). That is, we assume that for any set of \(k\) positive real numbers \(a_1, a_2, \dots, a_k\), the following inequality is true:

\[
\frac{a_1 + a_2 + \dots + a_k}{k} \geq \sqrt[k]{a_1 \cdot a_2 \cdot \dots \cdot a_k}
\]

This is our **inductive hypothesis**.

We need to prove that the inequality also holds for \(n = k+1\). That is, we need to show:

\[
\frac{a_1 + a_2 + \dots + a_{k+1}}{k+1} \geq \sqrt[k+1]{a_1 \cdot a_2 \cdot \dots \cdot a_{k+1}}
\]

#### Step 1: Group the First \(k\) Terms

We can group the first \(k\) terms and the \((k+1)\)-th term separately. From the inductive hypothesis, we know that:

\[
\frac{a_1 + a_2 + \dots + a_k}{k} \geq \sqrt[k]{a_1 \cdot a_2 \cdot \dots \cdot a_k}
\]

Now, we need to incorporate the term \(a_{k+1}\). To do this, apply the two-variable AM-GM inequality to the average of \(\frac{a_1 + a_2 + \dots + a_k}{k}\) and \(a_{k+1}\):

\[
\frac{\frac{a_1 + a_2 + \dots + a_k}{k} + a_{k+1}}{2} \geq \sqrt{\frac{a_1 + a_2 + \dots + a_k}{k} \cdot a_{k+1}}
\]

Multiplying both sides of this inequality by 2 gives:

\[
\frac{a_1 + a_2 + \dots + a_k + a_{k+1}}{k+1} \geq \sqrt[k+1]{a_1 \cdot a_2 \cdot \dots \cdot a_{k+1}}
\]

Thus, we have shown that the inequality holds for \(n = k+1\).

### Conclusion

By the principle of mathematical induction, the arithmetic mean is always greater than or equal to the geometric mean for any set of \(n\) positive real numbers. Thus, the inequality:

\[
\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot \dots \cdot a_n}
\]

holds for all \(n \geq 1\).

### Do you want more details on any step?

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**Relative Questions:**
1. What is the difference between the arithmetic mean and geometric mean?
2. How is the AM-GM inequality useful in optimization problems?
3. Can the AM-GM inequality be extended to more general means (like harmonic mean)?
4. How does the inequality behave if all numbers are equal?
5. Can you provide examples where the equality in AM-GM holds?

**Tip:** In many inequality proofs, the two-variable AM-GM inequality is a helpful starting point for induction.

Let a1, a2, ..., an be positive real numbers. Prove by mathematical induction that the arithmetic mean of these numbers is greater or equals to the geometric mean.

Learn how to prove the Arithmetic Mean - Geometric Mean (AM-GM) Inequality using mathematical induction. This proof covers core algebraic concepts such as arithmetic and geometric means, the AM-GM inequality, and the principle of mathematical induction. Suitable for advanced high school students and college learners.

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