Problem 132 - A Complete Resource Book in Mathematics for JEE Main

Question

Assertion: 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 less than 1 . Reason: A.M. >G.M. for unequal numbers

Step-by-Step Solution

Answer

Yes, the assertion is true due to the strict inequality derived using A.M.-G.M. and Cauchy-Schwarz inequalities.

Step 1: Understanding the Given Assertion

We are given that \(a, b, c\) are distinct positive real numbers and \(a^2 + b^2 + c^2 = 1\). We need to show that \(ab + bc + ca < 1\).

Step 2: Analyzing the Use of A.M.-G.M. Inequality

The arithmetic mean-geometric mean (A.M.-G.M.) inequality for three positive numbers \(a^2, b^2, c^2\) states that \(\frac{a^2 + b^2 + c^2}{3} \geq \sqrt[3]{a^2 b^2 c^2}\). Given that \(a^2 + b^2 + c^2 = 1\), this simplifies our situation.

Step 3: Applying A.M.-G.M. Inequality

Since \(a^2 + b^2 + c^2 = 1\), the inequality becomes \(\frac{1}{3} \geq \sqrt[3]{a^2 b^2 c^2}\). Thus, \(a^2b^2c^2 \leq \left(\frac{1}{3}\right)^3 \).

Step 4: Relating to the Expression \(ab + bc + ca\)

We want to show \(ab + bc + ca < 1\). Notice that \((ab + bc + ca)^2 \leq (a^2 + b^2 + c^2)^2\), which is from Cauchy-Schwarz inequality stating \((xy + yz + zx)^2 \leq (x^2 + y^2 + z^2)(1 + 1 + 1)\).

Step 5: Applying Cauchy-Schwarz Inequality

According to Cauchy-Schwarz, \((ab + bc + ca)^2 \leq (a^2 + b^2 + c^2)(b^2 + c^2 + a^2)\), which simplifies to \(ab + bc + ca \leq \sqrt{(a^2 + b^2 + c^2)(b^2 + c^2 + a^2)} = \sqrt{1 \cdot 1} = 1\).

Step 6: Distinct Positive Real Numbers

Since \(a, b, c\) are distinct positive real numbers, they cannot all be equal. This ensures that the application of A.M. > G.M. inequality indeed results in a strict inequality, i.e., \(ab + bc + ca < 1\).

Step 7: Conclusion

Therefore, by the application of these inequalities, we have demonstrated the assertion that \(ab + bc + ca < 1\) when \(a, b, c\) are distinct positive real numbers and \(a^2 + b^2 + c^2 = 1\).

Key Concepts

Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is a crucial mathematical concept that helps us compare sums and products in inequalities. It can be particularly useful when dealing with expressions involving several variables, like in this problem where we have three variables: \(a\), \(b\), and \(c\). The inequality states that for any real numbers \(x_1, x_2, \ldots, x_n\) and \(y_1, y_2, \ldots, y_n\), the following holds:

• \((x_1y_1 + x_2y_2 + \ldots + x_ny_n)^2 \leq (x_1^2 + x_2^2 + \ldots + x_n^2)(y_1^2 + y_2^2 + \ldots + y_n^2)\)

In the context of our exercise, this inequality allows us to deduce that the square of the sum of products \((ab + bc + ca)^2\) is less than or equal to the product of sums \((a^2 + b^2 + c^2)(b^2 + c^2 + a^2)\). Given that \(a^2 + b^2 + c^2 = 1\), this simplifies to exactly 1, providing us a useful upper bound.
Essentially, Cauchy-Schwarz gives us broad applicability in these types of problems, making it a fundamental tool in proving or deriving inequalities.

Positive Real Numbers

The exercise specifies that \(a\), \(b\), and \(c\) are distinct positive real numbers. Let's break down what each of these terms means.

• **Positive Numbers:** These are numbers greater than zero. This means \(a\), \(b\), and \(c\) must be greater than zero, excluding zero and negative values.
• **Real Numbers:** These include all the numbers on the number line, excluding imaginaries. From negative to positive, and everything in between - it's all part of the real number family. In this exercise, they are positive, so we focus only on numbers greater than zero.
• **Distinct Values:** This means each of the variables \(a\), \(b\), and \(c\) has a unique value, none of them are equal. Consequently, inequalities requiring non-equality of numbers, like A.M. > G.M., can strictly be applied.

The fact that \(a, b, c\) are positive real numbers not only ensures the validity of inequalities like A.M.-G.M. and Cauchy-Schwarz but also guarantees certain outcomes, like ensuring \(ab + bc + ca < 1\) when \(a^2 + b^2 + c^2 = 1\), by preventing equality among them.

Mathematical Inequality Proof

Proving mathematical inequalities often involves multiple steps and the application of various mathematical principles or pre-existing inequalities. In our exercise, one such inequality, the A.M.-G.M. inequality, was utilized to draw conclusions about the relationship between arithmetic means and geometric means.
The proof initiated by exploring the aggregate sum of squares set to 1: \(a^2 + b^2 + c^2 = 1\). From this, through application of the A.M.-G.M. inequality, we determined:

• \(\frac{a^2 + b^2 + c^2}{3} \geq \sqrt[3]{a^2b^2c^2}\)
• Since the total is 1, it leads to \(\frac{1}{3} \geq \sqrt[3]{a^2b^2c^2}\)

Armed with these relationships and utilizing the Cauchy-Schwarz inequality, we systematically demonstrate that \((ab + bc + ca)^2\) cannot exceed 1. In conclusion, detailed proofs often use structured steps to bridge known inequalities and properties of numbers effectively, as seen in establishing \(ab + bc + ca < 1\). These mathematical processes highlight how engaging with multiple strategies can solidly prove assertions regarding inequalities.