Question

Beweisen Sie die folgende Verallgemeinerung der BERNOULLIschen Ungleichung: $$ \left(1+a_{1}\right)\left(1+a_{2}\right) \cdots \cdot\left(1+a_{n}\right)>1+\left(a_{1}+a_{2}+\cdots+a_{n}\right) $$ für \(a_{k}>0\) und \(n \geq 2\).

Short answer

The inequality holds by induction for all \(n \geq 2\) and \(a_k > 0\).

Step-by-step solution

Understand the Inequality

The goal is to prove the inequality for all numbers \(a_k > 0\) and integers \(n \geq 2\). The inequality stated is: \( (1 + a_1)(1 + a_2) \cdots (1 + a_n) > 1 + (a_1 + a_2 + \cdots + a_n) \). This is a generalization of the Bernoulli's inequality.

Simplify the Left Side

Express the product on the left side: \((1+a_1)(1+a_2)\cdots(1+a_n)\) by expanding it to include every possible combination of \(a_i\) sums. The expansion does indeed include the sum \(a_1 + a_2 + \cdots + a_n\) as well as additional positive terms.

Compare Expansion with Right Side

The right side of the inequality is simply \(1 + (a_1 + a_2 + \cdots + a_n)\). Since \((1 + a_1)(1 + a_2) \cdots (1 + a_n)\) expands to contain this sum plus additional positive terms, by definition the product will always exceed \(1 + (a_1 + a_2 + \cdots + a_n)\).

Use Induction to Prove for All n

Perform a proof by induction on \(n\). For the base case where \(n = 2\), we have \((1 + a_1)(1 + a_2) = 1 + a_1 + a_2 + a_1a_2\), which is greater than \(1 + a_1 + a_2\). Assume the inequality holds for some \(k = n\). Then, for \(k = n+1\), \((1+a_1)(1+a_2)\cdots(1+a_n)(1+a_{n+1}) = (1+a_1+a_2+\cdots+a_n+a_1a_{n+1}+\cdots+a_ka_{n+1})\), which adds additional positive terms. Hence, it also holds for \(n+1\).

Conclude the Proof

The base case and induction step demonstrate that the inequality is true for all \(n \geq 2\). Therefore, for any \(a_k > 0\) and integer \(n \geq 2\), \((1+a_1)(1+a_2)\cdots(1+a_n) > 1+(a_1+a_2+\cdots+a_n)\) is proven.

Key concepts

Proof by Induction

Proof by induction is a powerful mathematical technique used to prove statements that are said to be true for all natural numbers. It involves two main steps: proving the base case and then demonstrating the induction step.

• Base Case: This involves showing that the statement holds for the initial value, usually when the variable equals 1 or another fixed number, like 2 in our problem.
• Induction Step: Once the base case is proven, the next step is to assume that the statement holds for some arbitrary case, say at step "k", then prove it for step "k + 1". This is often referred to as the inductive hypothesis.

This method works because if the initial case is true, and each step leads naturally to the next, then the statement must be true for all steps. In our problem, once we show it works for two factors, we use that result to prove it works for three by assuming it works for any "n" to prove it works for "n+1".

Generalization of Inequalities

The inequality in question is a generalization of Bernoulli's inequality. Generalizations are important in mathematics because they extend specific instances of inequalities to broader situations, offering comprehensive insights and utility.
In the original Bernoulli's inequality, you might see expressions like \( (1 + a)^n \geq 1 + na \) for \( a > -1 \) and \( n \) being a non-negative integer. Our task here goes beyond this by considering products of multiple terms like \( (1 + a_1)(1 + a_2) \cdots (1 + a_n) \).

• These products involve real numbers greater than zero, which allow for more generalized statements compared to the classic forms.
• Each pair of terms interacts combinatorially, ensuring the resulting expression on the left side is always larger.

By showing these generalized forms, mathematicians open new avenues for applying inequalities in various fields such as calculus, algebra, and real-world scenarios.

Mathematical Induction

Mathematical induction is a sophisticated tool in mathematics for proving statements, especially mathematical formulas and inequalities like the one we are handling. Here, the process of induction resembles a domino effect.
When demonstrating this technique, think of three primary components:

• Anchor or Base Case: Validate that the inequality holds true for a minimal starting point like \( n = 2 \).
• Assumption or Inductive Hypothesis: Assume it's true for some number \( n = k \).
• Induction Step or Inductive Conclusion: Use the truth of the \( n = k \) case to prove it works when \( n = k + 1 \), completing the cycle for all integers greater than or equal to the base case.

This technique efficiently checks statements over potentially infinite collections of numbers by linking smaller, more manageable assertions to broader claims.

Expansion of Products

When dealing with expressions like \( (1 + a_1)(1 + a_2) \cdots (1 + a_n) \), you naturally encounter the expansion of products. This involves expressing the product as a sum of terms that each contain one or more of the \( a_i \).
The process includes calculating all possible combinations of these parts while ensuring each term only consists of positive products and sums like:

• The single elements: \( a_1, a_2, \cdots, a_n \)
• Multi-element combinations: Such as \( a_1a_2, a_1a_3, \cdots \)

These expansions add layers of complexity that facilitate demonstrating inequalities since each additional term is positive, undeniably making the expanded expression larger than the direct summed expression addressed in the inequality's right side.
Breaking down these expressions into their component sums makes understanding and proving inequalities easier, lending clarity to complex algebraic statements.