Question 475152
The inequality is true iff


*[tex \LARGE (a^3 + b^3)^2 < (c^3)^2 = c^6]


*[tex \LARGE (a^3 + b^3)^2 < (a^2 + b^2)^3]


*[tex \LARGE a^6 + 2a^3b^3 + b^6 < a^6 + 3a^4b^2 + 3a^2b^4 + b^6]


The a^6 + b^6 on both side cancel out, so we want to show for positive a,b


*[tex \LARGE 2a^3b^3 < 3a^4b^2 + 3a^2b^4]


Divide both sides by (a^2)(b^2)


*[tex \LARGE 2ab < 3a^2 + 3b^2]


By AM-GM inequality, *[tex \LARGE \frac{3a^2 + 3b^2}{2} \ge \sqrt{9a^2b^2} = 3ab]


This implies that *[tex \LARGE 3a^2 + 3b^2 \ge 6ab > 2ab], so we're done.