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.

It is correct, but I would explain the "since $a\ge b$ then $a^3 - a^2 > b^3 - b^2$" step a bit more.
\begin{align*}a^2 - a > b^2-b &\iff a(a^2-a) > a(b^2-b) \quad \text{(since $a>0$)} \\ &\iff a(a^2-a)>a(b^2-b)\ge b(b^2-b) \quad \text{(since $b\le a$)} \\ &\iff a^3-a^2 > b^3 -b^2 \\ &\iff a^3 + b^3 > a^2 + b^2\end{align*}
