<s>[INST] How do I prove $a^3-b^3 \geq a^2b - b^2a$ Given that $a>b>0$, prove that $a^3-b^3 \geq a^2b - b^2a$.
I have considered difference of cubes, where $a^3-b^3 = (a-b)(a^2+ab+b^2)$. However this doesn't seem to get me that far - especially when working backwards from the statement I need to prove - where I factorised the right hand side into $ab(a-b)$ and equated with the expansion earlier, giving the following inequality: $a^2+ab+b^2 \geq ab$. However by AM-GM, it follows that $a^2+b^2 \geq 2ab$ which my statement above does not follow. Help would be much appreciated.
 [/INST] Given that $a > b > 0$, we take
$$
S = a^3 - b^3, \ \ T = a^2 b - b^2 a = a b (a - b)
$$
Note that we can write
$$
S = a^3 - b^3 = (a - b) (a^2 + a b + b^2)
$$
Thus,
$$
S - T = (a - b) (a^2 + a b + b^2) - a b (a - b) = (a - b)(a^2 + b^2) > 0
$$
since $a > b > 0$.
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