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

We are tasked with proving that if \( w^3 + x^3 + y^3 + z^3 = 10 \), then \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \).

### Step 1: Use of the Arithmetic Mean-Geometric Mean (AM-GM) Inequality

The **AM-GM inequality** for four variables \( w^3, x^3, y^3, z^3 \) states that:

\[
\frac{w^3 + x^3 + y^3 + z^3}{4} \geq \sqrt[4]{w^3 x^3 y^3 z^3}
\]

However, this inequality may not directly help in this context because it doesn't relate powers of 4. Instead, we need to manipulate the given equation and estimate \( w^4 + x^4 + y^4 + z^4 \).

### Step 2: Use Power Mean Inequality

The **Power Mean Inequality** tells us that for any set of non-negative numbers and exponents \( p > q \):

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

In our case, take \( p = 4 \) and \( q = 3 \), and the terms are \( w, x, y, z \). Applying this inequality, we get:

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

Since \( w^3 + x^3 + y^3 + z^3 = 10 \), we substitute this into the inequality:

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

Simplifying:

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

Now, raise both sides to the fourth power:

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

Multiply both sides by 4:

\[
w^4 + x^4 + y^4 + z^4 \geq 4 \cdot 2.5^{\frac{4}{3}}
\]

### Step 3: Approximate \( 2.5^{\frac{4}{3}} \)

Now, compute \( 2.5^{\frac{4}{3}} \):

\[
2.5^{\frac{4}{3}} \approx 3.937
\]

Thus:

\[
w^4 + x^4 + y^4 + z^4 \geq 4 \cdot 3.937 \approx 15.748
\]

### Step 4: Compare with \( \sqrt[3]{2500} \)

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

\[
\sqrt[3]{2500} \approx 13.58
\]

Since \( w^4 + x^4 + y^4 + z^4 \geq 15.748 \), we conclude that:

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

Thus, the inequality holds.

### Conclusion

We have shown that if \( w^3 + x^3 + y^3 + z^3 = 10 \), then \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \).

---

Would you like further details on any step or another related topic?

### 5 Related Questions:
1. What is the Power Mean Inequality, and how is it generally applied?
2. Can the AM-GM inequality be applied directly in this situation?
3. How does the Power Mean Inequality compare with the Hölder's inequality?
4. How do you calculate cube roots of large numbers manually?
5. How does the inequality change if the powers were different?

### Tip: 
When working with inequalities, approximating roots or exponents using a calculator can help simplify complex expressions.

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 > 3√2500 given w^3 + x^3 + y^3 + z^3 = 10. This problem explores key concepts in algebra and inequalities, utilizing the Power Mean Inequality and AM-GM Inequality. Suitable for students in Grades 11-12.

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