If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.

Question

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?

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score 3 · Answer 1 · answered Jul 07 '22 at 03:37

3

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} $

answered Jul 07 '22 at 03:37

Mingqiu Du

31

Great, but I wanted to ask that how did you come up with this solution? – Philosophical Nature Jul 07 '22 at 10:24

score 2 · Accepted Answer · edited Jul 07 '22 at 02:31

2

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}

edited Jul 07 '22 at 02:31

answered Jul 07 '22 at 02:25

Wang YeFei

6,390

Great, but I wanted to ask that how did you come up with this solution? – Philosophical Nature Jul 07 '22 at 10:24
First use $4ab < (a+b)^2$ to get an intequality in terms of $a+b$ and simply solve for $a+b$ – Wang YeFei Jul 07 '22 at 16:17

score 1 · Answer 3 · edited Jul 07 '22 at 04:07

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 $$ 3/4Q<1\equiv Q<4/3. $$

If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.

3 Answers3