<s>[INST] If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align*}
b^{3} - b^{2} = a^{3} - a^{2} & \Longleftrightarrow b^{3} - a^{3} = b^{2} - a^{2}\\\\
& \Longleftrightarrow (b-a)(a^{2} + ab + b^{2}) = (b+a)(b-a)\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = b + a
\end{align*}
but what next?
 [/INST] Rearrange $a^3-a^2=b^3-b^2$ to get
$$
b^3-a^3=b^2-a^2.
$$
Since $a\neq b$ we can factor out $(b-a)$ from both sides to get
$$
a^2+ab+b^2=a+b
$$
which is equivalent to
$$
(a+b)^2-ab=a+b
$$
Let $Q\equiv a+b$ then by the AM-GM inequality $\sqrt ab\leq(a+b)/2=Q/2$, and since $a\neq b$ the inequality is strict ($\sqrt ab < (a+b)/2$). Squaring this gives $ab<Q^2/4$.
We now have
$$
Q^2-Q<Q^2/4
$$
and since $a>0$, $b>0$ we can factor out a $Q$ and rearrange to get
$$
3/4Q<1\equiv Q<4/3.
$$
</s>