<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] Refer to 'Inequality of arithmetic and geometric means' in wiki
$\frac{{a + b}}{2} \geqslant \sqrt {ab}  \Rightarrow \frac{{{{(a + b)}^2}}}{4} \geqslant ab$
Next,
$\begin{gathered}
  {a^2} + ab + {b^2} = a + b \hfill \\
  {(a + b)^2} = a + b + ab < a + b + \frac{{{{(a + b)}^2}}}{4} \hfill \\
  \frac{{3{{(a + b)}^2}}}{4} < a + b \hfill \\
  a + b < \frac{4}{3} \hfill \\ 
\end{gathered} $
</s>