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