Problem 18 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a, b, c\) are distinct positive 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) any real number

Step-by-Step Solution

Answer

(A) less than 1

Step 1: Understand the Problem

We are given three distinct positive real numbers \(a, b, c\) such that their squares sum to 1: \(a^2 + b^2 + c^2 = 1\). We need to find the possible value(s) of \(ab + bc + ca\).

Step 2: Apply the AM-GM Inequality

Using the inequality \(a^2 + b^2 + c^2 \geq ab + bc + ca\), and knowing \(a^2 + b^2 + c^2 = 1\), we apply the inequality:\[1 = a^2 + b^2 + c^2 \geq ab + bc + ca\]This shows that \(ab + bc + ca\) is less than or equal to 1.

Step 3: Check Exact Equality Conditions

The equality \(a^2 + b^2 + c^2 = ab + bc + ca\) holds when \(a = b = c\), but since \(a, b, c\) are distinct, equality cannot occur. Therefore, \(ab + bc + ca\) must be strictly less than 1.

Key Concepts

AM-GM Inequality

The AM-GM Inequality, or Arithmetic Mean-Geometric Mean Inequality, is a fundamental principle in algebra. It states that for any list of non-negative real numbers, the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM). This is written mathematically as:

• \[\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 \cdot x_2 \cdots x_n}\]

For two numbers, this reduces to:

• \[\frac{a + b}{2} \geq \sqrt{ab}\]

This inequality is useful for various mathematical proofs and problem-solving, as demonstrated in the given exercise. In the problem, we use the AM-GM Inequality on the squares of three distinct real numbers \(a, b, \) and \(c\). We know:

• \[a^2 + b^2 + c^2 \geq ab + bc + ca\]

Which, given that \(a^2 + b^2 + c^2 = 1\), informs us that the sum \(ab + bc + ca\) must be less than or equal to this value, specifically it cannot reach one when \(a, b, \) and \(c\) are distinct.

Distinction of Real Numbers

In algebra, dealing with distinct real numbers is fundamental. When we say numbers are distinct, it means no two numbers are alike. In the exercise, we are told \(a, b,\) and \(c\) are distinct positive real numbers. This distinction implies certain conditions in equations cannot hold.
For example, if these numbers were not distinct, they might all be equal, breaking the inequality constraint from the previous section. The condition of being distinct is crucial when applying inequalities because certain properties or simplifications only apply when numbers are not exactly the same.
In our problem, because \(a = b = c\) does not apply, we satisfy the condition that the expression \(ab + bc + ca\) cannot reach equality in the AM-GM Inequality, reinforcing that it must be strictly less than 1.

Sum of Squares Identity

The sum of squares identity is an algebraic tool used to compute expressions involving squares of terms. The expression \(a^2 + b^2 + c^2\) is the sum of the squares of \(a, b,\) and \(c\). In many algebra problems, understanding the properties of these sums is key.
For the exercise in question, \(a^2 + b^2 + c^2 = 1\) serves as the main defining condition. This equation is used alongside known identities and inequalities to determine the bounds or properties of other expressions linked to \(a, b,\) and \(c\).
The relationship of the sum of squares with other forms, such as the product sums \(ab + bc + ca\), is crucial. Here, it is leveraged with the AM-GM Inequality to reason why \(ab + bc + ca\) cannot equal 1. When you see such sums or identities in mathematics, look for strategic ways to apply them to simplify problems or find confinements.