<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] Also, we can make the following.
Since by the condition $1>\frac{a+b}{a^2+b^2},$ by C-S we obtain:
$$a^3+b^3>\frac{(a^3+b^3)(a+b)}{a^2+b^2}\geq\frac{(a^2+b^2)^2}{a^2+b^2}=a^2+b^2.$$
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