Problem 12 - Algebra for JEE ADVANCED

Question

If \(a, b, c\) are the sides of a triangle, then the minimum value of \(\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\) is equal to a. 3 b. 6 c. 9 d. 12

Step-by-Step Solution

Answer

The minimum value is 3.

Step 1: Identify the expression

The expression we need to evaluate is \( \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \). This expression involves the sides of a triangle \(a, b, c\).

Step 2: Apply the triangle inequality

From the triangle inequality, we know that for any triangle with sides \(a, b, c\): \(a < b + c\), \(b < c + a\), and \(c < a + b\). This ensures all denominators are positive.

Step 3: Consider a symmetrical case

To simplify, consider \(a = b = c\). In this case, the expression becomes \(\frac{a}{a+a-a}+\frac{a}{a+a-a}+\frac{a}{a+a-a} = \frac{a}{a} + \frac{a}{a} + \frac{a}{a} = 3\).

Step 4: Evaluate using AM-GM inequality

Apply the AM-GM inequality to \( \frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c} \). According to AM-GM, the arithmetic mean is at least the geometric mean: \( \frac{\frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c}}{3} \geq \sqrt[3]{\frac{a}{b+c-a} \times \frac{b}{c+a-b} \times \frac{c}{a+b-c}} \). Calculating for each equal gives \(3\times \sqrt[3]{1}\).

Step 5: Conclusion about the minimum value

Both the symmetrical case and application of inequalities suggest that the minimum possible value of the expression is \(3\).

Key Concepts

Symmetrical Case

When evaluating expressions involving multiple variables, finding simplifications can often help. One common approach is considering symmetrical cases. This means assuming all variables are equal. In the context of the problem, the expression \( \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \) simplifies significantly for symmetrical cases. By setting \(a = b = c\), the expression becomes more manageable: \( \frac{a}{a} + \frac{a}{a} + \frac{a}{a} \). Thus, the expression simplifies to 3. This is visualized by:

• Each part of the expression simplifies to \( \frac{a}{a}\), which equals 1.
• Adding these parts gives \(1 + 1 + 1 = 3\).

Using symmetrical cases in problems helps identify patterns and often leads to meaningful results quickly. It also provides insights that may not be immediately apparent from the algebraic expression.

AM-GM Inequality

The AM-GM inequality is a powerful mathematical tool used to find relationships between means of numbers. It states that the arithmetic mean (AM) is always greater than or equal to the geometric mean (GM) for any set of positive numbers. This can be expressed as:

• \( \frac{x_1 + x_2 + ... + x_n}{n} \geq \sqrt[n]{x_1 x_2 ... x_n} \).

In the context of our triangle inequality problem, we apply AM-GM to show that:

• The arithmetic mean \( \frac{\frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c}}{3} \)
• Is at least equal to the geometric mean \( \sqrt[3]{\frac{a}{b+c-a} \times \frac{b}{c+a-b} \times \frac{c}{a+b-c}} \).

The application of AM-GM to symmetric and balanced cases often shows that this equality holds, contributing to finding the minimum value of expressions. In particular, when values are equal (as in our symmetrical case with \(a = b = c\)), the inequality transitions to an equality, further supporting the calculated minimum of 3.

Minimum Value of Expression

Finding the minimum value of the expression \( \frac{a}{b+c-a} + \frac{b}{c+a-b} + \frac{c}{a+b-c} \) involves understanding the underlying properties of triangles and inequalities. By leveraging both the symmetrical cases and the AM-GM inequality, we can deduce that 3 is the minimal achievable value under the given conditions. When the sides of the triangle are equal, the expression suggests a natural equilibrium that leads to the minimal expression sum. Here's why this makes sense:

• The denominators and numerators become straightforward when \(a = b = c\), simplifying the problem.
• Applying AM-GM shows this symmetry achieves the boundary (minimum) condition.

Triangular relationships ensure all terms stay positive and define the conditions under which the terms are summed optimally. Using both the symmetrical case and the AM-GM inequality thus points consistently toward a minimum expression value of 3, confirming the validity of using these approaches.