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

To prove that \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \), given the equation \( w^3 + x^3 + y^3 + z^3 = 10 \), let's analyze the problem step by step.

### Step 1: Cauchy-Schwarz Inequality

We can apply the **Cauchy-Schwarz inequality** in the following form:
\[
(w^4 + x^4 + y^4 + z^4) \cdot (1 + 1 + 1 + 1) \geq (w^3 + x^3 + y^3 + z^3)^2
\]
Simplifying, we get:
\[
4(w^4 + x^4 + y^4 + z^4) \geq (w^3 + x^3 + y^3 + z^3)^2
\]
Substitute \( w^3 + x^3 + y^3 + z^3 = 10 \) into the equation:
\[
4(w^4 + x^4 + y^4 + z^4) \geq 10^2
\]
\[
4(w^4 + x^4 + y^4 + z^4) \geq 100
\]
\[
w^4 + x^4 + y^4 + z^4 \geq \frac{100}{4} = 25
\]

### Step 2: Compare with \( \sqrt[3]{2500} \)

Next, compute \( \sqrt[3]{2500} \):
\[
\sqrt[3]{2500} \approx 13.58
\]
Thus, we need to show that:
\[
w^4 + x^4 + y^4 + z^4 \geq 25 > 13.58
\]

This inequality holds since \( 25 > \sqrt[3]{2500} \approx 13.58 \). Therefore, we have shown that:
\[
w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}
\]

### Conclusion

Hence, we have proved that:
\[
w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}
\]

Would you like more details or clarifications?

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Here are 5 related questions:
1. Can you apply the Cauchy-Schwarz inequality in higher dimensions?
2. How does the Cauchy-Schwarz inequality work in other norms, like \( L^p \)-spaces?
3. What are the properties of cubic roots in inequalities?
4. How would this proof change if \( w^3 + x^3 + y^3 + z^3 \neq 10 \)?
5. Can we generalize this result to more than four variables?

**Tip:** The Cauchy-Schwarz inequality is a powerful tool for dealing with sums of powers in inequalities.

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

Learn how to prove that w^4 + x^4 + y^4 + z^4 is greater than the cubic root of 2500 using the Cauchy-Schwarz inequality. This advanced algebra problem covers key mathematical concepts, including inequalities and cubic roots. Suitable for students in Grades 10-12.

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