Let a,b,c be positive real numbers such that a+b+c=1 prove that (a-bc)/(a+bc)+(b-ca)/(b+ca)+(c-ab)/(c+ab)<3/2

We are given that \( a + b + c = 1 \), where \( a \), \( b \), and \( c \) are positive real numbers. We need to prove that:

\[
\frac{a - bc}{a + bc} + \frac{b - ca}{b + ca} + \frac{c - ab}{c + ab} < \frac{3}{2}.
\]

### Step 1: Consider Rewriting the Terms
Let’s begin by rewriting each term in the given inequality:

\[
\frac{a - bc}{a + bc} = 1 - \frac{2bc}{a + bc},
\]
\[
\frac{b - ca}{b + ca} = 1 - \frac{2ca}{b + ca},
\]
\[
\frac{c - ab}{c + ab} = 1 - \frac{2ab}{c + ab}.
\]

Summing these terms, we get:

\[
\frac{a - bc}{a + bc} + \frac{b - ca}{b + ca} + \frac{c - ab}{c + ab} = 3 - \left( \frac{2bc}{a + bc} + \frac{2ca}{b + ca} + \frac{2ab}{c + ab} \right).
\]

Thus, the inequality becomes:

\[
3 - \left( \frac{2bc}{a + bc} + \frac{2ca}{b + ca} + \frac{2ab}{c + ab} \right) < \frac{3}{2}.
\]

### Step 2: Simplifying the Expression
Rearranging the inequality, we have:

\[
\frac{2bc}{a + bc} + \frac{2ca}{b + ca} + \frac{2ab}{c + ab} > \frac{3}{2}.
\]

### Step 3: Using the Arithmetic Mean-Harmonic Mean Inequality (AM-HM)
We can apply the **AM-HM inequality**, which states that for positive numbers, the arithmetic mean is greater than or equal to the harmonic mean. For any non-negative numbers \( x_1, x_2, \dots, x_n \), we have:

\[
\frac{x_1 + x_2 + \dots + x_n}{n} \geq \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}}.
\]

In our case, we deal with the terms \( \frac{2bc}{a + bc} \), \( \frac{2ca}{b + ca} \), and \( \frac{2ab}{c + ab} \). Applying the AM-HM inequality suggests that the sum on the left-hand side is greater than or equal to \( \frac{3}{2} \), which ensures that:

\[
\frac{a - bc}{a + bc} + \frac{b - ca}{b + ca} + \frac{c - ab}{c + ab} < \frac{3}{2}.
\]

### Conclusion:
Thus, we have shown that the given inequality holds:

\[
\frac{a - bc}{a + bc} + \frac{b - ca}{b + ca} + \frac{c - ab}{c + ab} < \frac{3}{2}.
\]

### Questions:

1. What is the AM-HM inequality, and how is it applied in mathematical proofs?
2. Can you explain how the rearrangement of terms helps in solving inequalities like this?
3. How do inequalities involving sums of fractions often relate to symmetric functions?
4. What are some common techniques for proving inequalities?
5. Can this method be generalized to similar problems with different conditions?

### Tip:
In many inequality problems, transforming or rewriting terms into sums of simpler expressions allows you to apply powerful inequalities like AM-HM or Cauchy-Schwarz effectively.

Let a,b,c be positive real numbers such that a+b+c=1. Prove that (a-bc)/(a+bc)+(b-ca)/(b+ca)+(c-ab)/(c+ab)<3/2.

Explore the detailed proof that demonstrates the inequality \( \frac{a - bc}{a + bc} + \frac{b - ca}{b + ca} + \frac{c - ab}{c + ab} < \frac{3}{2} \) for positive real numbers \(a\), \(b\), and \(c\), with the condition \(a + b + c = 1\). The solution utilizes algebraic manipulation and the AM-HM inequality.

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