Problem 9 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(a, b, c\) are distinct +ve real numbers and \(a^{2}+b^{2}+c^{2}=1\) then \(a b+b c+c a\) is (a) less than 1 (b) equal to 1 (c) greater than 1 (d) anyreal no.

Step-by-Step Solution

Answer

The expression \(ab + bc + ca\) is less than 1 (option a).

Step 1: Understand the given conditions

We have three distinct positive real numbers: \(a, b, c\) and the equation \(a^2 + b^2 + c^2 = 1\). We need to analyze \(ab + bc + ca\).

Step 2: Use the AM-GM inequality

According to the AM-GM (Arithmetic Mean - Geometric Mean) inequality for non-negative numbers, \(ab + bc + ca \leq \frac{(a+b+c)^2}{3}\). We need to further explore this inequality in the context of the given condition.

Step 3: Express \((a+b+c)^2\) in terms of \(a^2, b^2, c^2, ab, bc, ca\)

Expand \((a+b+c)^2\) to get \(a^2 + b^2 + c^2 + 2(ab + bc + ca)\). We know from the given condition that \(a^2 + b^2 + c^2 = 1\), so substituting this, we have:\((a+b+c)^2 = 1 + 2(ab + bc + ca)\).

Step 4: Use the identity and AM-GM to find bounds

Since \((a+b+c)^2\) is the square of a sum, it is non-negative. Thus,1. If \((a+b+c)^2 = 1 + 2(ab + bc + ca)\), and using AM-GM inequality for 1 variable:\(ab + bc + ca < a^2 + b^2 + c^2 = 1\). Therefore \(ab + bc + ca < 1\).

Step 5: Conclude the values of \(ab + bc + ca\)

Since \(a, b, c\) are positive and distinct, by the inequality derived, \(ab + bc + ca < 1\). This means the correct choice is (a): less than 1.

Key Concepts

AM-GM Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) inequality is a fundamental principle in algebra. It applies to a set of non-negative real numbers. This inequality states that the arithmetic mean (average) of the numbers is always greater than or equal to their geometric mean. In mathematical terms:

• For any non-negative real numbers, say \(x_1, x_2, \, \ldots, \, x_n \), the inequality is given by: \[ \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 \cdot x_2 \cdot \cdots \cdot x_n} \]
• Even though it is a simple concept, its implications are vast and can be used to solve a range of problems.
• The equality condition of the AM-GM inequality holds only when all the numbers involved are equal.

In our context, we are dealing with the expressions involving \(a, b, c\). The AM-GM inequality helps us determine that \(ab + bc + ca \leq \frac{(a+b+c)^2}{3} \). This is because the three terms \(ab\), \(bc\), and \(ca\) form a set of non-negative real numbers. Understanding how to apply this inequality is crucial for solving equations involving sums and products of numbers.

Real Numbers

Real numbers comprise a wide range of numbers including positive, negative, and zero values. They are the building blocks of almost all algebraic expressions. Real numbers can be either rational, like integers and fractions, or irrational, like the square root of non-perfect squares or numbers like \(\pi\).

• Positive real numbers are those that are greater than zero.
• In algebra, the properties of real numbers are used to manipulate and analyze expressions and equations.
• Various operations such as addition, subtraction, multiplication, and division (except by zero) are well-defined within real numbers, making them versatile in solving numerous mathematical problems.

In our exercise, \(a, b, c\) are not only real numbers but **distinct** and **positive** ones. This distinction is important as it bears on the application of the AM-GM inequality, confirming uniqueness and positivity in calculations.

Algebraic Identities

Algebraic identities are equations that hold true for all values of the variables involved. They help simplify complex algebraic expressions. An example is the identity of the square of a sum:

• \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)

In the problem under analysis, this identity helps in transforming the equation to evaluate \(ab + bc + ca\). By substituting the given value \(a^2 + b^2 + c^2 = 1\) into the expanded form, we effectively leverage algebraic identities to derive useful conclusions:

• We obtained: \((a+b+c)^2 = 1 + 2(ab + bc + ca)\)
• It allows us to analyze and set bounds; showing algebraic identities provide essential tools in simplifying and solving equations.

These identities enable us to calculate the sought expressions more efficiently and play a critical role in proofs and problem-solving across all levels of mathematics.