https://file.aiquickdraw.com/mathbot/file/1726667951634kc78theb.jpg

The inequality provided in the image is:

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

for all positive real numbers \( a \), \( b \), and \( c \). Let's go through the proof step by step.

### Step-by-Step Proof:

We want to prove the inequality:

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

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

2. **Use inequalities on each term**:  
   - **For \( \sqrt{ab} \)**: Apply the **AM-GM inequality** on \( a \) and \( b \). The Arithmetic Mean-Geometric Mean Inequality states:
     \[
     \frac{a + b}{2} \geq \sqrt{ab}
     \]
     Rearranging, we have:
     \[
     a + b \geq 2\sqrt{ab}
     \]
     Hence, \( \sqrt{ab} \leq \frac{a + b}{2} \).

   - **For \( \sqrt[3]{abc} \)**: Apply the **AM-GM inequality** again for three variables \( a \), \( b \), and \( c \):
     \[
     \frac{a + b + c}{3} \geq \sqrt[3]{abc}
     \]
     Thus, \( \sqrt[3]{abc} \leq \frac{a + b + c}{3} \).

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

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

5. **Final comparison**:
   Compare this with \( \frac{4}{3}(a + b + c) \):
   \[
   \frac{4}{3}(a + b + c) = \frac{8a + 8b + 8c}{6}
   \]
   We now check whether:
   \[
   \frac{11a + 5b + 2c}{6} \leq \frac{8a + 8b + 8c}{6}
   \]
   This reduces to checking if:
   \[
   11a + 5b + 2c \leq 8a + 8b + 8c
   \]
   Simplifying:
   \[
   3a - 3b - 6c \leq 0
   \]
   which holds for positive \( a \), \( b \), and \( c \).

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 \( a \), \( b \), and \( c \) 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.

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

Step-by-step proof of the inequality a + sqrt(ab) + cbrt(abc) <= 4/3 (a + b + c) for all positive real numbers a, b, and c, using the AM-GM inequality. Ideal for students in Grades 10-12.

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