<s>[INST] If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ we get $a^2-a \gt b-b^2$
Since $a \geq b$ we can get $a^3-a^2 \gt b^2-b^3$ $\Rightarrow$ $a^3+b^3 \gt a^2+b^2$
If this solution is incorrect, please explain why and attach the correct solution. Thank you.
 [/INST] I think the step when you arrive at $a^3-a^2>b^2-b^3$ is incorrect as you multiply inequalities whose expressions can be negative.
Alternatively, note that $a^3+a\ge 2a^2$ (this is equivalent to $a(a-1)^2 \ge 0$) and analogously $b^3+b\ge 2b^2$. Therefore
$$a^3+b^3+a+b\ge 2(a^2+b^2) > a^2+b^2+a+b,$$
where the second inequality uses the assumption $a^2+b^2>a+b$. As a consequence, $a^3+b^3>a^2+b^2$.
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