Math Problem Statement

a, b, c are positive numbers, prove that it is not possible for the inequalitiesa(1−b)>1/4,b(1−c)>1/4,c(1−a)>1/4, toholdatthesametime

Solution

To prove that the inequalities a(1b)>14a(1 - b) > \frac{1}{4}, b(1c)>14b(1 - c) > \frac{1}{4}, and c(1a)>14c(1 - a) > \frac{1}{4} cannot all hold simultaneously, we will assume that they do and derive a contradiction.

Given:

  1. a(1b)>14a(1 - b) > \frac{1}{4}
  2. b(1c)>14b(1 - c) > \frac{1}{4}
  3. c(1a)>14c(1 - a) > \frac{1}{4}

First, sum these inequalities: a(1b)+b(1c)+c(1a)>14+14+14=34a(1 - b) + b(1 - c) + c(1 - a) > \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}

Expanding each term on the left side: aab+bbc+cca>34a - ab + b - bc + c - ca > \frac{3}{4}

Rearrange the terms: (a+b+c)(ab+bc+ca)>34(a + b + c) - (ab + bc + ca) > \frac{3}{4}

Since a,b,ca, b, c are positive numbers, we use the AM-GM inequality on the sum ab+bc+caab + bc + ca: ab+bc+ca(a+b+c)23ab + bc + ca \leq \frac{(a + b + c)^2}{3}

Thus, (a+b+c)(a+b+c)23>34(a + b + c) - \frac{(a + b + c)^2}{3} > \frac{3}{4}

Let x=a+b+cx = a + b + c. Substituting xx into the inequality, we get: xx23>34x - \frac{x^2}{3} > \frac{3}{4}

Multiply through by 3 to clear the fraction: 3xx2>943x - x^2 > \frac{9}{4}

Rewrite the inequality: x23x+94<0x^2 - 3x + \frac{9}{4} < 0

To solve this quadratic inequality, first consider the equality: x23x+94=0x^2 - 3x + \frac{9}{4} = 0

The discriminant (Δ\Delta) of this quadratic equation is: Δ=b24ac=(3)24194=99=0\Delta = b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot \frac{9}{4} = 9 - 9 = 0

Since the discriminant is zero, the quadratic equation has a double root at: x=32x = \frac{3}{2}

This means the quadratic x23x+94x^2 - 3x + \frac{9}{4} touches the x-axis at x=32x = \frac{3}{2} and does not cross it, so x23x+940x^2 - 3x + \frac{9}{4} \geq 0.

Therefore, for any positive xx, x23x+940x^2 - 3x + \frac{9}{4} \geq 0, implying: x23x+94<0 is impossible.x^2 - 3x + \frac{9}{4} < 0 \text{ is impossible.}

This contradiction implies that our initial assumption that all three inequalities hold simultaneously is incorrect. Hence, it is not possible for the inequalities a(1b)>14a(1 - b) > \frac{1}{4}, b(1c)>14b(1 - c) > \frac{1}{4}, and c(1a)>14c(1 - a) > \frac{1}{4} to all hold at the same time.

Would you like more details or have any questions?

Here are five questions you might want to ask next:

  1. How do you apply the AM-GM inequality to other types of problems?
  2. Can you explain the discriminant and its significance in solving quadratic equations?
  3. What are some other common inequalities used in mathematical proofs?
  4. How can we visualize the solution to a quadratic inequality?
  5. Can you provide examples of contradictions in other mathematical proofs?

Tip: When solving inequalities involving multiple variables, considering the sum or product of the inequalities can often help to find contradictions or simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Inequalities
AM-GM Inequality
Quadratic Equations

Formulas

AM-GM inequality
Quadratic discriminant

Theorems

AM-GM inequality
Quadratic discriminant

Suitable Grade Level

Advanced High School