If w^3+x^3+y^3+z^3=10 show that w^4+y^4+x^4+z^4>^3√2500

We are given the equation \( w^3 + x^3 + y^3 + z^3 = 10 \), and we need to show that:

\[
w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}
\]

### Step-by-Step Approach:

1. **Apply the Power Mean Inequality**:

The Power Mean Inequality states that for non-negative real numbers \( a_1, a_2, \dots, a_n \) and exponents \( p < q \), we have:

\[
\left( \frac{a_1^p + a_2^p + \dots + a_n^p}{n} \right)^{\frac{1}{p}} \leq \left( \frac{a_1^q + a_2^q + \dots + a_n^q}{n} \right)^{\frac{1}{q}}
\]

For our case:
- Set \( a_1 = w \), \( a_2 = x \), \( a_3 = y \), and \( a_4 = z \).
- Take \( p = 3 \) and \( q = 4 \).

This gives us:

\[
\left( \frac{w^3 + x^3 + y^3 + z^3}{4} \right)^{\frac{1}{3}} \leq \left( \frac{w^4 + x^4 + y^4 + z^4}{4} \right)^{\frac{1}{4}}
\]

Substitute \( w^3 + x^3 + y^3 + z^3 = 10 \):

\[
\left( \frac{10}{4} \right)^{\frac{1}{3}} \leq \left( \frac{w^4 + x^4 + y^4 + z^4}{4} \right)^{\frac{1}{4}}
\]

Simplifying:

\[
\left( 2.5 \right)^{\frac{1}{3}} \leq \left( \frac{w^4 + x^4 + y^4 + z^4}{4} \right)^{\frac{1}{4}}
\]

Cube both sides:

\[
2.5 \leq \frac{w^4 + x^4 + y^4 + z^4}{4}
\]

Multiply by 4:

\[
10 \leq w^4 + x^4 + y^4 + z^4
\]

2. **Compare to \( \sqrt[3]{2500} \)**:

Now, we compare \( w^4 + x^4 + y^4 + z^4 \) with \( \sqrt[3]{2500} \).

We compute \( \sqrt[3]{2500} \):

\[
\sqrt[3]{2500} = 13.58 \text{ (approximately)}
\]

Since \( w^4 + x^4 + y^4 + z^4 \geq 10 \), and \( 10 < 13.58 \), we have:

\[
w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}
\]

Thus, the inequality holds.

### Conclusion:
We have shown that \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \).

Do you need further clarification on any steps?

Here are some related questions to explore:
1. What is the Power Mean Inequality and how is it derived?
2. How does the Power Mean Inequality apply to different exponents?
3. How would the inequality change if we had \( w^3 + x^3 + y^3 + z^3 = 20 \)?
4. Can the Cauchy-Schwarz inequality be used to solve similar problems?
5. How does the result change if we cube both sides instead of using the Power Mean Inequality?

**Tip**: The Power Mean Inequality is a very useful tool when comparing different powers of sums, especially in inequalities involving higher-order terms.

If w^3 + x^3 + y^3 + z^3 = 10, show that w^4 + y^4 + x^4 + z^4 > ^3√2500

Explore how to prove the inequality w^4 + y^4 + x^4 + z^4 > ^3√2500 given that w^3 + x^3 + y^3 + z^3 = 10. This problem delves into algebraic concepts and the application of the Power Mean Inequality, suitable for advanced high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation