Least Negative Most Positive?

Hi,

Firstly, we can deduce that

$\begin{cases} \alpha +\beta =7-\gamma \\ \alpha \beta =-{50}/{\gamma } \\ \left| \alpha \right| +\left| \beta \right| +\left| \gamma \right| =b \\ \left| \alpha \beta \gamma \right| =d=50\quad \end{cases}$

WLOG, assume that $\gamma < 0 < \alpha \le \beta$

$\Rightarrow b=\alpha +\beta -\gamma =7-2\gamma$

The problem now can be interpreted as finding the minimum integer value of $-2\gamma$ .

$\alpha ,\beta ,\gamma \in R$ requires that the quadratic equation ${ X }^{ 2 }-(\alpha +\beta )X+\alpha \beta =0$ or ${ X }^{ 2 }-(7-\gamma )X- { 50 }/{ \gamma } =0$ has real root(s) for some real value of $\gamma$ .

$\Leftrightarrow { (7-\gamma ) }^{ 2 }\ge -200/\gamma$

$\Rightarrow { (-2\gamma ) }^{ 3 }+28{ (-2\gamma ) }^{ 2 }+196(-2\gamma )-1600\ge 0 \quad (1)$ (note that $\gamma < 0$ )

We use a little calculus here.

Consider the function $f\left( -2\gamma \right) =LHS\Rightarrow \begin{cases} f^{ ' }\left( -14 \right) =0 \\ f^{ ' }\left( -14/3 \right) =0 \\ f(-14)f(-14/3)>0 \end{cases}\Rightarrow \exists !\delta \in R:f(\delta )=0$

Even more, we know that $f\left( 4 \right) f\left( 5 \right) <0\Rightarrow 4<\delta <5$ .

Therefore, $(1) \Leftrightarrow -2\gamma \ge \delta$ . What we need here is the smallest integer that is greater than or equal to $\delta$ .

Hence, $-2 \gamma=\left\lceil \delta \right\rceil =5$ .

In conclusion, the minimum "integral" value of $b+d$ is $62$ at $(\alpha, \beta, \gamma) = (u, v, -5/2)$ and its permutations in which $u, v$ are roots of ${ y }^{ 2 }-9.5y+20=0$

The word "integral" makes me so confused.

d=50 Now apply AM>=GM FOR THE ROOTS OF 2 CUBIC As d = 50 its minimum value is 11.05. But this is minimum hence minimum integer value is 12 . Hence answer= 62