Team Selection Tests

Let $x_1, x_2, \dots, x_{24}$ be the roots of the polynomial $P(x)$. Then $0 < x_1, x_2, \dots, x_{24} \le \alpha$, $x_1x_2 \cdots x_{24} = 1$ and we have to prove
$$|(x_1 - 1)(x_2 - 1) \cdots (x_{24} - 1)| \le \left(\frac{19}{5}\right)^5 (\alpha - 1)^{24}.$$
If among the numbers $x_1, x_2, \dots, x_{24}$ there is a number equal to $1$ then the inequality is obviously true. Consider the case where these numbers are all different from $1$. Without loss of generality, assume
$$x_1 \le x_2 \le \cdots \le x_k < 1 < x_{k+1} \le x_{k+2} \le \cdots \le x_{24} \le \alpha.$$
The inequality to be proven is rewritten as
$$(1 - x_1)(1 - x_2) \cdots (1 - x_k)(x_{k+1} - 1)(x_{k+2} - 1) \cdots (x_{24} - 1) \le \left(\frac{19}{5}\right)^5 (\alpha - 1)^{24}. \quad (1)$$
Let $t = \sqrt[k]{x_1x_2\cdots x_k}$. Using the AM-GM inequality, we have
$$(1 - x_1)(1 - x_2) \cdots (1 - x_k) \le \left( \frac{k - x_1 - x_2 - \cdots - x_k}{k} \right)^k \le (1 - t)^k. \quad (2)$$
Now, for each $k+1 \le i \le 24$, let $x_i = e^{y_i}$ and let $l = e^{\frac{y_{k+1}+\cdots+y_{24}}{24-k}}$. Obviously $0 < y_{k+1} \le \cdots \le y_{24} \le \ln \alpha$. Consider the function $h(x) = \ln(e^x - 1)$ on the domain $(0, +\infty)$, we have $h''(x) = -\frac{e^x}{(e^x-1)^2} < 0$ so the function $h$ is concave on the domain $(0, +\infty)$. Therefore, according to Jensen's inequality, we have
$$h(y_{k+1}) + \cdots + h(y_{24}) \le (24-k)h\left(\frac{y_{k+1} + \cdots + y_{24}}{24-k}\right), \text{ or } \\ (x_{k+1} - 1)(x_{k+2} - 1) \cdots (x_{24} - 1) \le (l-1)^{24-k}. \quad (3)$$
Using the inequalities (2) and (3), it can be seen that, to prove the inequality (1), we only need to prove
$$(1 - t)^k (l - 1)^{24-k} \le \left(\frac{19}{5}\right)^5 (\alpha - 1)^{24}.$$
Since $t^k l^{24-k} = 1$ then $t = l^{-\frac{24-k}{k}}$ the above inequality can be rewritten as
$$\left(1 - \frac{1}{l^{\frac{24-k}{k}}}\right)^k (l-1)^{24-k} \le \left(\frac{19}{5}\right)^5 (\alpha-1)^{24}. \quad (4)$$
Since $1 < l \le \alpha$ we have
$$\begin{align*} \text{LHS of (4)} &= \left(l^{\frac{24-k}{k}} - 1\right)^k \left(1 - \frac{1}{l}\right)^{24-k} \\ &\le \left(\alpha^{\frac{24-k}{k}} - 1\right)^k \left(1 - \frac{1}{\alpha}\right)^{24-k} = \frac{\left(\alpha^{\frac{24-k}{k}} - 1\right)^k (\alpha - 1)^{24-k}}{\alpha^{24-k}} \end{align*}$$
Thus, to prove the inequality (4), we only need to prove
$$\frac{\left(\alpha^{\frac{24-k}{k}} - 1\right)^k (\alpha - 1)^{24-k}}{\alpha^{24-k}} \le \left(\frac{19}{5}\right)^5 (\alpha - 1)^{24}$$
or
$$f(\alpha) \le \left(\frac{19}{5}\right)^5, \quad (5)$$
where
$$f(x) = \frac{\left(x^{\frac{24-k}{k}} - 1\right)^k}{x^{24-k} (x-1)^k}.$$
We have $\ln f(x) = k \ln \left(x^{\frac{24-k}{k}} - 1\right) - (24-k) \ln x - k \ln(x-1)$.
Therefore
$$\frac{f'(x)}{f(x)} = \frac{(24-k)x^{\frac{24-k}{k}}}{x\left(x^{\frac{24-k}{k}} - 1\right)} - \frac{24-k}{x} - \frac{k}{x-1} = - \frac{kx^{\frac{24}{k}} - 24x + (24-k)}{x(x-1)\left(x^{\frac{24-k}{k}} - 1\right)}$$
Using the AM-GM inequality, we have $kx^{\frac{24}{k}} - 24x + (24-k) > 0$. Therefore $f'(x) < 0$, implies $f$ is a decreasing function on the domain $(1, +\infty)$. Therefore, we deduce that
$$f(\alpha) \le \lim_{x \to 1^+} f(x) = \left(\frac{24-k}{k}\right)^k.$$
Thus, to prove the inequality (5), we only need to prove that
$$\left(\frac{24-k}{k}\right)^k \le \left(\frac{19}{5}\right)^5,$$
or
$$\ln(24-k) - \ln k \le \frac{5}{k} \ln \frac{19}{5}.$$
Consider the function $g(x) = \ln(24-x) - \ln x - \frac{5}{x} \ln \frac{19}{5}$ with $1 \le x \le 23$. We have
$$g'(x) = -\frac{1}{24-x} - \frac{1}{x} + \frac{5}{x^2} \ln \frac{19}{5} = -\frac{x(5 \ln \frac{19}{5} + 24) - 120 \ln \frac{19}{5}}{x^2(24-x)}$$
The equation $g'(x) = 0$ has a unique solution $x_0$. In addition, it is easy to see that $g'(5)g'(6) < 0$ so $x_0 \in (5, 6)$. Because $g'(x) > 0$ with $x < x_0$ and $g'(x) < 0$ with $x > x_0$, $g(x)$ reaches its maximum value at $x = x_0$. Because $k$ is a positive integer, we have $g(k) \le \max\{g(5), g(6)\} = 0$. $\square$