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

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

---

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.

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

Explore how to prove that w^4 + x^4 + y^4 + z^4 exceeds the cube root of 2500, given that w^3 + x^3 + y^3 + z^3 = 10. This detailed solution uses Jensen's inequality and convex functions to provide a step-by-step explanation, suitable for 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