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.

Your proof is correct. Indeed:
$$\begin{cases}a^2-a \gt b-b^2\\
a\ge b>0\end{cases} \Rightarrow \\
a(a^2-a)>b(b-b^2) \Rightarrow \\
a^3-a^2>b^2-b^3 \Rightarrow \\
a^3+b^3>a^2+b^2.$$
Alternative proof. Consider $b=ax, x\ge 1$. Then:
$$a^2+b^2 \gt a+b \Rightarrow \\
a^2+a^2x^2>a+ax \stackrel{\text{divide by} \ a}{\Rightarrow} \\
a+ax^2>1+x \Rightarrow \\
a>\frac{1+x}{1+x^2} \ \ (1)$$
Hence:
$$a^3+b^3>a^2+b^2 \iff \\
a^3+a^3x^3>a^2+a^2x^2 \Rightarrow \\
a+ax^3>1+x^2 \iff \\
a(1+x^3)\stackrel{(1)}{>}\frac{1+x}{1+x^2}(1+x^3)\ge 1+x^2 \iff \\
(1+x)(1+x^3)\ge(1+x^2)^2 \iff \\
1+x+x^3+x^4\ge1+2x^2+x^4 \iff \\
x(1+x^2)\ge 2x^2 \iff \\
(x-1)^2\ge 0.$$
It is the Cauchy-Swarz inequality (given by Michael Rozenberg).
