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The problem in the image states:

**If \(x > 0\) and \(y > 0\), prove that:**
\[
2x^3 + 2y^3 \geq (x^2 + y^2)(x + y)
\]

### Solution:
We need to prove the inequality:
\[
2x^3 + 2y^3 \geq (x^2 + y^2)(x + y)
\]

Start by expanding the right-hand side:
\[
(x^2 + y^2)(x + y) = x^2(x + y) + y^2(x + y) = x^3 + x^2y + y^3 + y^2x
\]
Thus, the inequality becomes:
\[
2x^3 + 2y^3 \geq x^3 + x^2y + y^3 + y^2x
\]

Now, subtract \(x^3 + y^3\) from both sides:
\[
x^3 + y^3 \geq x^2y + y^2x
\]

We aim to show that:
\[
x^3 + y^3 \geq x^2y + y^2x
\]
This can be rewritten as:
\[
x^3 - x^2y \geq y^2x - y^3
\]
or equivalently:
\[
x^2(x - y) \geq y^2(x - y)
\]

Since \(x > 0\) and \(y > 0\), we can factor out \((x - y)\) from both sides:
\[
(x^2 - y^2)(x - y) \geq 0
\]

This is true because \(x^2 \geq y^2\) and hence:
\[
x^3 + y^3 \geq x^2y + y^2x
\]
This completes the proof.

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Do you want further details or explanations?

Here are 5 related questions:
1. What does it mean for an inequality to be "true for all \(x > 0\) and \(y > 0\)"?
2. Can you prove similar inequalities for higher powers, such as \(x^4 + y^4\)?
3. How does the inequality change if \(x\) and \(y\) are negative?
4. What role does symmetry play in this inequality?
5. What are other common inequalities involving cubes and squares?

**Tip:** Always start by simplifying both sides of an inequality to see if they can be factored or written in a more comparable form.

If x > 0 and y > 0, prove that: 2x^3 + 2y^3 ≥ (x^2 + y^2)(x + y)

Step-by-step solution to proving the inequality 2x^3 + 2y^3 ≥ (x^2 + y^2)(x + y). Covers algebraic concepts, expanding cubic expressions, and factorization techniques. Ideal for students in Grades 10-12.

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