To prove that $(a^2 + b^2)(a^4 + b^4) > a^3 + b^3$ for real numbers $a$ and $b$, let's analyze both sides of the inequality.

Step 1: Expanding the Left Side

The left side is $(a^2 + b^2)(a^4 + b^4)$, so let's expand it: $(a^2 + b^2)(a^4 + b^4) = a^6 + a^2b^4 + b^2a^4 + b^6.$

So, the left side becomes: $a^6 + b^6 + a^2b^4 + b^2a^4.$

Step 2: Examining Each Term

Each term in this expression is non-negative for real $a$ and $b$. Also, $a^6$ and $b^6$ are typically larger than $a^3$ and $b^3$, respectively, especially as $a$ and $b$ grow in magnitude.

Step 3: Comparing Terms

Now let's consider the expression on the right side, $a^3 + b^3$, and compare it with each term in the expanded left side.

Notice:

1. $a^6 + b^6 \geq a^3 + b^3$, because $a^6 \geq a^3$ and $b^6 \geq b^3$ for real $a$ and $b$.
2. The additional terms $a^2b^4$ and $b^2a^4$ on the left side are both non-negative, which makes the left side strictly larger than the right side when both $a$ and $b$ are nonzero.

Conclusion

Thus, we conclude that: $(a^2 + b^2)(a^4 + b^4) > a^3 + b^3$ for real values of $a$ and $b$.

Would you like a deeper breakdown of any part of this proof or have any questions?

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Related Questions

1. How would this inequality change if we only considered positive values of $a$ and $b$?
2. What is the relationship between $a^6$ and $a^3$ for values of $a$ in the interval $0 < a < 1$?
3. How would this inequality behave if $a$ and $b$ were complex numbers?
4. Could we generalize this inequality to higher powers of $a$ and $b$?
5. What role does symmetry in $a$ and $b$ play in this proof?

Tip

When analyzing inequalities, expanding terms and comparing each part separately can clarify which side is larger, especially when one side has additional non-negative terms.