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?

We have:
\begin{align}
a^2+ab+b^2 = a+b &\implies 4(a+b)^2 - 4(a+b) = 4ab < (a+b)^2\\
&\implies 3(a+b)^2 - 4(a+b) < 0\\
&\implies (a+b)(3(a+b)-4) < 0\\
&\implies 3(a+b) - 4 < 0\\
&\implies (a+b) < \dfrac{4}{3}.
\end{align}
