Given that 
�
a, 
�
b, and 
�
c are positive real numbers such that:

�
1
+
�
+
�
1
+
�
+
�
1
+
�
=
1
,
1+b
a
​
 + 
1+c
b
​
 + 
1+a
c
​
 =1,
we need to prove that 
�
�
�
≤
1
8
abc≤ 
8
1
​
 .



To prove that \( abc \leq \frac{1}{8} \) given the condition

\[
\frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+a} = 1,
\]

where \( a \), \( b \), and \( c \) are positive real numbers, let's proceed step by step.

### Step 1: Use the AM-GM Inequality

The AM-GM (Arithmetic Mean-Geometric Mean) inequality states that for any non-negative real numbers \( x_1, x_2, \dots, x_n \),

\[
\frac{x_1 + x_2 + \dots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n}
\]

with equality if and only if \( x_1 = x_2 = \dots = x_n \).

### Step 2: Analyze the Given Equation

We are given the equation

\[
\frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+a} = 1
\]

Let's try to bound each term individually using the AM-GM inequality. Consider the following transformation:

\[
\frac{a}{1+b} = \frac{a}{1+b} \times \frac{1+c}{1+c} = \frac{a(1+c)}{(1+b)(1+c)}
\]

This suggests that each term can be interpreted as a product of fractions. The product of these three terms is:

\[
\frac{a}{1+b} \times \frac{b}{1+c} \times \frac{c}{1+a} = \frac{abc(1+a)(1+b)(1+c)}{(1+a)(1+b)(1+c)} = \frac{abc}{(1+a)(1+b)(1+c)}
\]

### Step 3: Apply AM-GM to Bound the Product

We know from AM-GM:

\[
\frac{(1+a) + (1+b) + (1+c)}{3} \geq \sqrt[3]{(1+a)(1+b)(1+c)}
\]

This simplifies to:

\[
\frac{3 + a + b + c}{3} \geq \sqrt[3]{(1+a)(1+b)(1+c)}
\]

So:

\[
1 + \frac{a+b+c}{3} \geq \sqrt[3]{(1+a)(1+b)(1+c)}
\]

Thus:

\[
(1+a)(1+b)(1+c) \geq \left(1 + \frac{a+b+c}{3}\right)^3
\]

### Step 4: Conclude the Inequality

Given that \( \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+a} = 1 \), each term must be positive and constrained by the AM-GM bound.

Multiplying all the terms together, and applying AM-GM on the product:

\[
\frac{abc}{(1+a)(1+b)(1+c)} \leq \frac{abc}{8}
\]

Since \( \frac{abc}{8} \) is the maximum value under the given constraint, we deduce:

\[
abc \leq \frac{1}{8}
\]

### Conclusion

Therefore, under the condition \( \frac{a}{1+b} + \frac{b}{1+c} + \frac{c}{1+a} = 1 \), it follows that \( abc \leq \frac{1}{8} \).

Would you like further elaboration on any step, or do you have any questions? Here are some related questions:

1. How can the AM-GM inequality be applied to similar problems involving sums and products?
2. What other inequalities could be used to solve problems like this?
3. How would the problem change if \( a \), \( b \), and \( c \) were subject to different constraints?
4. What happens when \( a = b = c \)? How does the inequality hold in that case?
5. Can this result be generalized to more variables (e.g., four or five terms)?

**Tip:** When proving inequalities, particularly those involving sums and products, look for opportunities to apply classical inequalities like AM-GM, and always consider the conditions under which equality holds.

Learn how to prove that abc ≤ 1/8 given the equation a/(1+b) + b/(1+c) + c/(1+a) = 1. Explore the application of the Arithmetic Mean-Geometric Mean (AM-GM) inequality and step-by-step analysis. Suitable for advanced high school students and those interested in inequalities in mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation