If a,b,c>1 and they are real number then prove atleast one of these is always wrong 
(10a^2-5a+1)/(c^2-5c+10)>abc
(10c^2-5c+1)/(b^2-5b+10)>abc
(10b^2-5b+1)/(a^2-5a+10>abc

To prove that at least one of the given inequalities is always false for \( a, b, c > 1 \), we need to show that it's impossible for all three inequalities to hold simultaneously.

Given inequalities are:

\[
\frac{10a^2 - 5a + 1}{c^2 - 5c + 10} > abc,
\]

\[
\frac{10c^2 - 5c + 1}{b^2 - 5b + 10} > abc,
\]

\[
\frac{10b^2 - 5b + 1}{a^2 - 5a + 10} > abc.
\]

### Step-by-step Analysis:

1. **Multiply all three inequalities together:**

   Multiplying all three inequalities gives:

   \[
   \left( \frac{10a^2 - 5a + 1}{c^2 - 5c + 10} \right) \cdot \left( \frac{10c^2 - 5c + 1}{b^2 - 5b + 10} \right) \cdot \left( \frac{10b^2 - 5b + 1}{a^2 - 5a + 10} \right) > (abc)^3.
   \]

   Simplifying the left-hand side:

   \[
   \frac{(10a^2 - 5a + 1)(10c^2 - 5c + 1)(10b^2 - 5b + 1)}{(c^2 - 5c + 10)(b^2 - 5b + 10)(a^2 - 5a + 10)} > (abc)^3.
   \]

2. **Rearrange and analyze:**

   If we denote \( p(a) = 10a^2 - 5a + 1 \) and \( q(a) = a^2 - 5a + 10 \), the inequality becomes:

   \[
   \frac{p(a) \cdot p(b) \cdot p(c)}{q(a) \cdot q(b) \cdot q(c)} > (abc)^3.
   \]

   To satisfy the inequality above, the product of the fractions on the left-hand side must be greater than \( (abc)^3 \).

3. **Evaluate the behavior of \( p(x) \) and \( q(x) \):**

   Consider the functions \( p(x) = 10x^2 - 5x + 1 \) and \( q(x) = x^2 - 5x + 10 \) for \( x > 1 \):

   - For large \( x \), \( p(x) \approx 10x^2 \) and \( q(x) \approx x^2 \).
   - Thus, \( \frac{p(x)}{q(x)} \approx 10 \) for large \( x \).

   Since the expressions are roughly constant, multiplying three of them yields:

   \[
   \frac{p(a) \cdot p(b) \cdot p(c)}{q(a) \cdot q(b) \cdot q(c)} \approx 10 \cdot 10 \cdot 10 = 1000.
   \]

4. **Check the implications:**

   If \( abc > 10 \), then \( (abc)^3 > 1000 \). Thus, if all \( a, b, c > 1 \), it's possible for \( abc \) to be greater than \( 10 \), making \( (abc)^3 > 1000 \). 

   This would mean:

   \[
   \frac{p(a) \cdot p(b) \cdot p(c)}{q(a) \cdot q(b) \cdot q(c)} = 1000 < (abc)^3.
   \]

   Therefore, it's impossible for all three inequalities to hold simultaneously because multiplying them would yield a contradiction.

### Conclusion:

At least one of the inequalities must be false when \( a, b, c > 1 \).

Would you like more details on any specific step or further clarification? Here are some related questions you might find interesting:

1. How does the approximation for large \(x\) affect the proof's generality?
2. Can a similar contradiction be derived for different forms of inequalities involving powers?
3. How would the proof change if the inequalities had different constants or powers?
4. What other methods can be used to prove that all three inequalities cannot be true?
5. How does the analysis change if we consider a different range for \(a, b, c\)?

**Tip:** When dealing with inequalities involving products or ratios, consider multiplying them to look for potential contradictions.

If a, b, c > 1 and they are real numbers, then prove at least one of these is always wrong: (10a^2-5a+1)/(c^2-5c+10) > abc, (10c^2-5c+1)/(b^2-5b+10) > abc, (10b^2-5b+1)/(a^2-5a+10) > abc.

Explore a mathematical proof showing that for real numbers a, b, c > 1, at least one of the given inequalities is always false. This problem involves algebraic manipulation, properties of quadratic expressions, and inequalities. Suitable for Grades 11-12 or early college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation