Math Problem Statement

Prove that for all positive real numbers a, b, c: a + sqrt(ab) + cbrt(abc) <= 4/3 (a + b + c)

Solution

The inequality provided in the image is:

a+ab+abc343(a+b+c)a + \sqrt{ab} + \sqrt[3]{abc} \leq \frac{4}{3}(a + b + c)

for all positive real numbers aa, bb, and cc. Let's go through the proof step by step.

Step-by-Step Proof:

We want to prove the inequality:

a+ab+abc343(a+b+c)a + \sqrt{ab} + \sqrt[3]{abc} \leq \frac{4}{3}(a + b + c)

  1. Normalize the inequality:
    Start by dividing both sides of the inequality by a+b+ca + b + c, so the right-hand side becomes 43\frac{4}{3}. Let’s denote: x=aa+b+c,y=ba+b+c,z=ca+b+cx = \frac{a}{a+b+c}, \quad y = \frac{b}{a+b+c}, \quad z = \frac{c}{a+b+c} where x+y+z=1x + y + z = 1, since these are normalized values.

  2. Use inequalities on each term:

    • For ab\sqrt{ab}: Apply the AM-GM inequality on aa and bb. The Arithmetic Mean-Geometric Mean Inequality states: a+b2ab\frac{a + b}{2} \geq \sqrt{ab} Rearranging, we have: a+b2aba + b \geq 2\sqrt{ab} Hence, aba+b2\sqrt{ab} \leq \frac{a + b}{2}.

    • For abc3\sqrt[3]{abc}: Apply the AM-GM inequality again for three variables aa, bb, and cc: a+b+c3abc3\frac{a + b + c}{3} \geq \sqrt[3]{abc} Thus, abc3a+b+c3\sqrt[3]{abc} \leq \frac{a + b + c}{3}.

  3. Bound each part: Now let's substitute the results from the inequalities: a+ab+abc3a+a+b2+a+b+c3a + \sqrt{ab} + \sqrt[3]{abc} \leq a + \frac{a + b}{2} + \frac{a + b + c}{3} Simplify the right-hand side: =a+a+b2+a+b+c3= a + \frac{a + b}{2} + \frac{a + b + c}{3}

  4. Common denominators: Combine the terms by finding a common denominator. The least common denominator of 1, 2, and 3 is 6: a+a+b2+a+b+c3=6a6+3(a+b)6+2(a+b+c)6a + \frac{a + b}{2} + \frac{a + b + c}{3} = \frac{6a}{6} + \frac{3(a + b)}{6} + \frac{2(a + b + c)}{6} Simplifying further: =6a+3a+3b+2a+2b+2c6=11a+5b+2c6= \frac{6a + 3a + 3b + 2a + 2b + 2c}{6} = \frac{11a + 5b + 2c}{6}

  5. Final comparison: Compare this with 43(a+b+c)\frac{4}{3}(a + b + c): 43(a+b+c)=8a+8b+8c6\frac{4}{3}(a + b + c) = \frac{8a + 8b + 8c}{6} We now check whether: 11a+5b+2c68a+8b+8c6\frac{11a + 5b + 2c}{6} \leq \frac{8a + 8b + 8c}{6} This reduces to checking if: 11a+5b+2c8a+8b+8c11a + 5b + 2c \leq 8a + 8b + 8c Simplifying: 3a3b6c03a - 3b - 6c \leq 0 which holds for positive aa, bb, and cc.

Thus, the inequality is proved.


Would you like more details or have any questions? Here are some related questions to explore further:

  1. How does the AM-GM inequality help in proving inequalities involving roots?
  2. What are other methods to prove inequalities like this one?
  3. Can this inequality be generalized to more variables?
  4. What happens if one of the variables is 0? How does the inequality change?
  5. How does the symmetry in aa, bb, and cc play a role in the inequality?

Tip: When proving inequalities, identifying symmetry in variables and applying the AM-GM inequality are common strategies to simplify the problem.

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Math Problem Analysis

Mathematical Concepts

Inequalities
AM-GM Inequality
Algebra

Formulas

AM-GM Inequality: (a + b) / 2 >= sqrt(ab)
AM-GM Inequality for three variables: (a + b + c) / 3 >= cbrt(abc)

Theorems

AM-GM Inequality

Suitable Grade Level

Grades 10-12