If $m(α)$ is global minimum of $ \ln(x^2 + αx + α)$, find $α$ such that $m(α)$ is maximal

Question

For any α in the interval (0, 4) let $m(α)$ be the global minimum of the function $$\ln(x^2 + αx + α)$$

For what value of α does $m(α)$ attains its maximum?

I'd prefer if the solution was as simple as possible, thanks.

Answer 1

Note

$$x^2+ax+a = \left(x+\frac a2\right)^2 + a - \frac {a^2}4 \ge a - \frac {a^2}4$$

Given that the function $\ln(t)$ is monotonically increasing, the global minimum of $\ln(x^2+ax+a)$ is at $a - \frac {a^2}4$, i.e.

$$m(a) = \ln \left(a - \frac {a^2}4 \right)$$

Note again,

$$a - \frac {a^2}4 = 1-\left(\frac a2-1\right)^2 \le 1$$

Thus, the maximum of $m(a)$ is attained at $a = 2$, i.e.

$$m_{max}(a) = m(2)=0$$