Maximize the sum $a+b$

Question

If the equation $x^4-4x^3+4x^2+ax+b=0$(where $a,b$ are reals) has $2$ distinct positive roots $\alpha, \beta$ such that $\alpha+\beta =2\alpha \beta$, then find the maximum value of $a+b$.

I have no idea how do I start using the given condition. Also, it appears to me that POSITIVE is an important word here. Some hints please. Thanks.

Answer 1

$\alpha,\beta$ are roots of a polynomial $$\tag1x^2-2sx+s$$ for some $s$ with $s>0$ (as $\alpha,\beta>0$) and $s\ne 2$ (as $\alpha\ne \beta$) and $D=4s^2-4s>0$ (to have two real solutions). This amounts to $s\in(1,2)\cup (2,\infty)$.

Polynomial division of the given polynomial by $(1)$ gives us a remainder of $$ \tag2(a + 8s^3 - 20s^2 + 12s)\cdot x + (b -4s^3 + 9s^2 - 4s)$$ and this must be zero. Hence $$a+b = -4s^3+11s^2-8s. $$ The term on the right has derivative $-12s^2+22s-8$, which has roots at $s=\frac12$ and $s=\frac43$. Only the second of these gives us a local maximum, and in fact this is a global maximum relative to the the domain $(1,2)\cup(2,\infty)$. By plugging in $s=\frac43$ we find $$a+b\le -\frac{16}{27}$$ and in fact by looking at $(2)$ coeficientwise, this bound is achieved precisely for $$a=\frac{16}{27},\quad b=-\frac{32}{27} $$ and in this case we have (up to order) $$ \alpha=2,\quad\beta=\frac23.$$

Answer 2

akech's solution is wrong at second part because the critical point $x=\dfrac{2}{3}$ of $x(x-2)^2$ is not the min point,it is local max point. You can't get result in this way even the final answer is same. here is right approach.

$a+b=-\dfrac{1}{2}(\alpha(\alpha-2)^2+\beta(\beta-2)^2)$

so we need to find min value of $(\alpha(\alpha-2)^2+\beta(\beta-2)^2)$

$t=\alpha+\beta=2\alpha\beta>0,(\alpha+\beta)^2\ge 4\alpha\beta \implies t^2\ge 2t \implies t \ge 2 \implies t>2 (\alpha \neq\beta)$

$\alpha(\alpha-2)^2+\beta(\beta-2)^2=(\alpha^3+\beta^3)-4(\alpha^2 +\beta^2)+4(\alpha+\beta)=(\alpha+\beta)(\alpha^2 +\beta^2-\alpha\beta)-4((\alpha+\beta))^2-2(\alpha\beta))+4t=t^3-\dfrac{11t^2}{2}+8t=f(t)$

$f'(t)=3t^2-11t+8=0 \implies t= \dfrac{8}{3}>2, t=1<2$

it is easy to verify $t= \dfrac{8}{3}$ is min point.