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?
 A: 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} $
A: 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}
A: 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
$$
3/4Q<1\equiv Q<4/3.
$$
