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

To prove that \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \) given that \( w^3 + x^3 + y^3 + z^3 = 10 \), we can apply the **Cauchy-Schwarz inequality**. Let's walk through the process step by step.

### 1. Applying Cauchy-Schwarz Inequality

The **Cauchy-Schwarz inequality** in its general form is:

\[
(a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + b_2^2 + \cdots + b_n^2) \geq (a_1 b_1 + a_2 b_2 + \cdots + a_n b_n)^2
\]

In this case, let's apply it to two sequences:
- \( a = (w^{2}, x^{2}, y^{2}, z^{2}) \)
- \( b = (1, 1, 1, 1) \)

This results in:

\[
(w^4 + x^4 + y^4 + z^4)(1^2 + 1^2 + 1^2 + 1^2) \geq (w^2 + x^2 + y^2 + z^2)^2
\]

Simplifying, we get:

\[
4(w^4 + x^4 + y^4 + z^4) \geq (w^2 + x^2 + y^2 + z^2)^2
\]

Thus:

\[
w^4 + x^4 + y^4 + z^4 \geq \frac{(w^2 + x^2 + y^2 + z^2)^2}{4}
\]

### 2. Estimating \( w^2 + x^2 + y^2 + z^2 \)

To find a bound for \( w^2 + x^2 + y^2 + z^2 \), let's use the fact that \( w^3 + x^3 + y^3 + z^3 = 10 \). By applying **Jensen's inequality** (because the function \( f(t) = t^{2/3} \) is concave), we know that:

\[
\frac{w^2 + x^2 + y^2 + z^2}{4} \geq \left( \frac{w^3 + x^3 + y^3 + z^3}{4} \right)^{2/3}
\]

Since \( w^3 + x^3 + y^3 + z^3 = 10 \), we have:

\[
\frac{w^2 + x^2 + y^2 + z^2}{4} \geq \left( \frac{10}{4} \right)^{2/3} = \left( 2.5 \right)^{2/3}
\]

Approximating the cube root:

\[
\left( 2.5 \right)^{2/3} \approx 1.737
\]

Thus:

\[
w^2 + x^2 + y^2 + z^2 \geq 4 \times 1.737 \approx 6.948
\]

### 3. Substituting into the Inequality

Now, substitute \( w^2 + x^2 + y^2 + z^2 \geq 6.948 \) into the inequality for \( w^4 + x^4 + y^4 + z^4 \):

\[
w^4 + x^4 + y^4 + z^4 \geq \frac{(6.948)^2}{4}
\]

\[
w^4 + x^4 + y^4 + z^4 \geq \frac{48.268}{4} \approx 12.067
\]

### 4. Comparing with \( \sqrt[3]{2500} \)

Next, calculate \( \sqrt[3]{2500} \):

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

Since \( 12.067 < 13.58 \), the inequality \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \) is **not directly satisfied**.

### 5. Adjusting the Bound

To show that \( w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500} \), we would need to improve our approximation or apply another approach (perhaps using more specific properties of the values \( w, x, y, z \)).

### Conclusion

We have demonstrated that \( w^4 + x^4 + y^4 + z^4 \geq 12.067 \), which is close but not greater than \( \sqrt[3]{2500} \). This indicates that a more refined approach or tighter bound is needed to complete the proof.

Would you like more details or clarifications on any steps?

Here are 5 follow-up questions:
1. How does Jensen's inequality work in bounding sums of powers?
2. Can the Cauchy-Schwarz inequality be used with different sequences for a tighter bound?
3. What is the significance of concavity in applying inequalities like Jensen's?
4. How can we improve estimates for cube roots in such problems?
5. Can an alternate inequality, like Hölder's, provide a better result?

**Tip:** When dealing with inequalities, always check if the functions involved are convex or concave—this determines which inequality (e.g., Jensen's) is applicable.

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

Explore the application of the Cauchy-Schwarz and Jensen's Inequalities to prove that w^4 + x^4 + y^4 + z^4 > ^3√2500, given that w^3 + x^3 + y^3 + z^3 = 10. This problem covers key advanced algebraic concepts and inequalities. Ideal for college-level students.

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