We are given the equation $w^3 + x^3 + y^3 + z^3 = 10$, and we are asked to show that $w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}$.

Step 1: Jensen's Inequality

We will use Jensen's Inequality for convex functions to help bound the expression $w^4 + x^4 + y^4 + z^4$. First, observe that the function $f(t) = t^4$ is convex because its second derivative is positive for all $t$ (since $f''(t) = 12t^2$).

Step 2: Apply Jensen’s Inequality

For convex functions, Jensen's inequality states that for non-negative numbers $a, b, c, d$, we have: $f\left( \frac{a + b + c + d}{4} \right) \leq \frac{f(a) + f(b) + f(c) + f(d)}{4}.$ In our case, applying this to the function $f(t) = t^4$, we get: $\left( \frac{w^3 + x^3 + y^3 + z^3}{4} \right)^4 \leq \frac{w^4 + x^4 + y^4 + z^4}{4}.$ Since we are given that $w^3 + x^3 + y^3 + z^3 = 10$, this becomes: $\left( \frac{10}{4} \right)^4 \leq \frac{w^4 + x^4 + y^4 + z^4}{4},$ which simplifies to: $\left( 2.5 \right)^4 \leq \frac{w^4 + x^4 + y^4 + z^4}{4}.$

Step 3: Simplifying

Now calculate $2.5^4$: $2.5^4 = (2.5 \times 2.5) \times (2.5 \times 2.5) = 6.25 \times 6.25 = 39.0625.$ Thus, we have: $39.0625 \leq \frac{w^4 + x^4 + y^4 + z^4}{4}.$ Multiplying both sides by 4, we get: $156.25 \leq w^4 + x^4 + y^4 + z^4.$

Step 4: Comparing with $\sqrt[3]{2500}$

Now we compute $\sqrt[3]{2500}$. We know that $2500^{1/3}$ is approximately: $\sqrt[3]{2500} \approx 13.58.$

Finally, observe that: $w^4 + x^4 + y^4 + z^4 \geq 156.25 > 13.58,$ which shows that: $w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}.$

Thus, we have shown that $w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}$.

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Would you like a more detailed explanation of any step or have any specific questions?

Here are 5 related questions to deepen your understanding:

1. What are the conditions under which Jensen’s inequality can be applied?
2. How do we calculate the cube root of a number like 2500 more accurately?
3. What makes a function convex, and why is this important for inequalities?
4. Can this method be applied to other powers, such as $w^5 + x^5 + y^5 + z^5$?
5. How does the choice of $w^3 + x^3 + y^3 + z^3 = 10$ affect the final inequality?

Tip: Convex functions grow faster than linear ones, which is why Jensen’s inequality helps to bound sums of powers effectively.