Math Problem Statement

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

Solution

We are given the equation w3+x3+y3+z3=10w^3 + x^3 + y^3 + z^3 = 10, and we need to show that:

w4+x4+y4+z4>25003w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}

Step-by-Step Approach:

  1. Apply the Power Mean Inequality:

The Power Mean Inequality states that for non-negative real numbers a1,a2,,ana_1, a_2, \dots, a_n and exponents p<qp < q, we have:

(a1p+a2p++anpn)1p(a1q+a2q++anqn)1q\left( \frac{a_1^p + a_2^p + \dots + a_n^p}{n} \right)^{\frac{1}{p}} \leq \left( \frac{a_1^q + a_2^q + \dots + a_n^q}{n} \right)^{\frac{1}{q}}

For our case:

  • Set a1=wa_1 = w, a2=xa_2 = x, a3=ya_3 = y, and a4=za_4 = z.
  • Take p=3p = 3 and q=4q = 4.

This gives us:

(w3+x3+y3+z34)13(w4+x4+y4+z44)14\left( \frac{w^3 + x^3 + y^3 + z^3}{4} \right)^{\frac{1}{3}} \leq \left( \frac{w^4 + x^4 + y^4 + z^4}{4} \right)^{\frac{1}{4}}

Substitute w3+x3+y3+z3=10w^3 + x^3 + y^3 + z^3 = 10:

(104)13(w4+x4+y4+z44)14\left( \frac{10}{4} \right)^{\frac{1}{3}} \leq \left( \frac{w^4 + x^4 + y^4 + z^4}{4} \right)^{\frac{1}{4}}

Simplifying:

(2.5)13(w4+x4+y4+z44)14\left( 2.5 \right)^{\frac{1}{3}} \leq \left( \frac{w^4 + x^4 + y^4 + z^4}{4} \right)^{\frac{1}{4}}

Cube both sides:

2.5w4+x4+y4+z442.5 \leq \frac{w^4 + x^4 + y^4 + z^4}{4}

Multiply by 4:

10w4+x4+y4+z410 \leq w^4 + x^4 + y^4 + z^4

  1. Compare to 25003\sqrt[3]{2500}:

Now, we compare w4+x4+y4+z4w^4 + x^4 + y^4 + z^4 with 25003\sqrt[3]{2500}.

We compute 25003\sqrt[3]{2500}:

25003=13.58 (approximately)\sqrt[3]{2500} = 13.58 \text{ (approximately)}

Since w4+x4+y4+z410w^4 + x^4 + y^4 + z^4 \geq 10, and 10<13.5810 < 13.58, we have:

w4+x4+y4+z4>25003w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}

Thus, the inequality holds.

Conclusion:

We have shown that w4+x4+y4+z4>25003w^4 + x^4 + y^4 + z^4 > \sqrt[3]{2500}.

Do you need further clarification on any steps?

Here are some related questions to explore:

  1. What is the Power Mean Inequality and how is it derived?
  2. How does the Power Mean Inequality apply to different exponents?
  3. How would the inequality change if we had w3+x3+y3+z3=20w^3 + x^3 + y^3 + z^3 = 20?
  4. Can the Cauchy-Schwarz inequality be used to solve similar problems?
  5. How does the result change if we cube both sides instead of using the Power Mean Inequality?

Tip: The Power Mean Inequality is a very useful tool when comparing different powers of sums, especially in inequalities involving higher-order terms.

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Math Problem Analysis

Mathematical Concepts

Algebra
Inequalities
Power Mean Inequality

Formulas

Power Mean Inequality: (a1^p + a2^p + ... + an^p)/n ≤ (a1^q + a2^q + ... + an^q)/n

Theorems

Power Mean Inequality

Suitable Grade Level

Grades 11-12