Problem 61 - University Calculus: Early Transcendentals

Question

An inequality for positive integers Show that if \(a, b, c,\) and \(d\) are positive integers, then $$\frac{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)\left(d^{2}+1\right)}{a b c d} \geq 16$$

Step-by-Step Solution

Answer

The inequality is true because \((a^2+1) \geq 2a\) for each variable.

Step 1: Express the Left Side

The inequality we want to prove is \( \frac{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}{abcd} \geq 16 \). We need to simplify and manipulate the left side to show it is greater than or equal to 16.

Step 2: Apply AM-GM Inequality

To use the AM-GM inequality on each term separately, note: \( a^2 + 1 \geq 2a \), \( b^2 + 1 \geq 2b \), \( c^2 + 1 \geq 2c \), \( d^2 + 1 \geq 2d \). You can justify these by rearranging: \( a^2 - 2a + 1 \geq 0 \), which is \((a-1)^2 \geq 0\), true for all integers.

Step 3: Multiply The Inequalities

Now multiply the inequalities obtained in Step 2: \((a^2+1)(b^2+1)(c^2+1)(d^2+1) \geq (2a)(2b)(2c)(2d) = 16abcd.\)

Step 4: Compare with the Original Expression

Now divide both sides by \( abcd \):\[\frac{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}{abcd} \geq \frac{16abcd}{abcd} = 16.\]This shows that the given inequality is indeed true for positive integers \(a, b, c,\) and \(d\).

Key Concepts

AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a powerful tool in mathematics, especially when comparing different types of means. It states that for any set of positive numbers, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM). In simple terms:

• If you have numbers like a_1, a_2, ..., a_n, the inequality can be expressed as \[\frac{a_1 + a_2 + ... + a_n}{n} \geq \sqrt[n]{a_1 \cdot a_2 \cdot ... \cdot a_n}.\]

This gets more interesting with equal numbers; the AM equals GM. To see the AM-GM inequality at play, consider the exercise where it's applied to each term separately:

• For \(a^2 + 1\), the inequality simplifies into \(a^2 + 1 \geq 2a\). This involves turning the original number into a friendly format that fits the AM-GM context.
• Similar forms exist for \(b, c,\) and \(d\) too, making it versatile.

That's how the AM-GM inequality helps simplify and solve inequalities like the one given in the exercise.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying mathematical expressions to reveal hidden properties or reach a desired result. In this problem, manipulating the given expression involves breaking it down and simplifying step by step.

• First, express each part individually. Simplify terms like \(a^2 + 1\) to show they're always positive.
• Use the AM-GM inequality as a stepping stone to replace terms with more manageable forms, like turning \(a^2+1\) into \(2a\).
• Once each term is simplified, multiply the resulting inequalities to form a more comprehensible expression.

By following these steps, the expression becomes easier to interpret and solve. This manipulation strategy is key in proving inequalities.

Positive Integers

In mathematics, the term positive integers refers to the set of numbers that are greater than zero and do not have any fractional or decimal part. These numbers, like 1, 2, 3, and so on, are essential in many mathematical proofs and exercises involving inequalities.

• Positive integers ensure the integrity of solutions; they allow inequalities and mathematical rules to consistently apply.
• In the exercise, knowing that \(a, b, c,\) and \(d\) are positive integers simplifies the expression \((a-1)^2 \geq 0\), making it straightforward to verify.
• Since positive integers are simple and whole, they support the straightforward application of concepts like AM-GM Inequality.

This ensures that all manipulations and products in the inequality yield valid, predictable results. Therefore, positive integers are the cornerstones for safely applying many mathematical tactics and strategies.