Team selection tests

Let $x_1,x_2,\dots ,x_{24}$ be the roots of the polynomial $P(x)$. Then $0<x_1,x_2,\dots ,x_{24}\le \alpha$, $x_1{x}_2\cdots x_{24}=1$ 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 $$(1-x_1)(1-x_2)\cdots (1-x_k)(x_{k+1}-1)(x_{k+2}-1)\cdots (x_{24}-1)\le {\left(\frac{19}{5}\right)}^5(\alpha -1{)}^{24}.(1)$$ Let $t=\sqrt[k]{x_1{x}_2\cdots x_k}$. Using the AM-GM inequality, we have $$(1-x_1)(1-x_2)\cdots (1-x_k)\le {\left(\frac{k-x_1-x_2-\cdots -x_k}{k}\right)}^k\le (1-t{)}^k.(2)$$ 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 $h^{\prime \prime }(x)=-\frac{e^x}{(e^x-1{)}^2}<0$ so the function $h$ is concave on the domain $(0,+\infty )$. Therefore, according to Jensen's inequality, we have $$\begin{gathered} h(y_{k+1})+\cdots +h(y_{24})\le (24-k)h\left(\frac{y_{k+1}+\cdots +y_{24}}{24-k}\right),\text{ or } \\ (x_{k+1}-1)(x_{k+2}-1)\cdots (x_{24}-1)\le (l-1{)}^{24-k}.(3) \end{gathered}$$ 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 $${\left(1-\frac{1}{l^{\frac{24-k}{k}}}\right)}^k(l-1{)}^{24-k}\le {\left(\frac{19}{5}\right)}^5(\alpha -1{)}^{24}.(4)$$ Since $1<l\le \alpha$ we have $$\begin{array}{rl}\text{LHS of (4)}& ={\left(l^{\frac{24-k}{k}}-1\right)}^k{\left(1-\frac{1}{l}\right)}^{24-k}\\ & \le {\left({\alpha }^{\frac{24-k}{k}}-1\right)}^k{\left(1-\frac{1}{\alpha }\right)}^{24-k}=\frac{{\left({\alpha }^{\frac{24-k}{k}}-1\right)}^k(\alpha -1{)}^{24-k}}{{\alpha }^{24-k}}\end{array}$$ 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 $$f(\alpha )\le {\left(\frac{19}{5}\right)}^5,(5)$$ 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 $$\frac{f^{\prime }(x)}{f(x)}=\frac{(24-k)x^{\frac{24-k}{k}}}{x\left(x^{\frac{24-k}{k}}-1\right)}-\frac{24-k}{x}-\frac{k}{x-1}=-\frac{k{x}^{\frac{24}{k}}-24x+(24-k)}{x(x-1)\left(x^{\frac{24-k}{k}}-1\right)}$$ Using the AM-GM inequality, we have $k{x}^{\frac{24}{k}}-24x+(24-k)>0$. Therefore $f^{\prime }(x)<0$, implies $f$ is a decreasing function on the domain $(1,+\infty )$. Therefore, we deduce that $$f(\alpha )\le \underset{x\to 1^{+}}{\lim }f(x)={\left(\frac{24-k}{k}\right)}^k.$$ 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 $$g^{\prime }(x)=-\frac{1}{24-x}-\frac{1}{x}+\frac{5}{x^2}\ln \frac{19}{5}=-\frac{x(5\ln \frac{19}{5}+24)-120\ln \frac{19}{5}}{x^2(24-x)}$$ The equation $g^{\prime }(x)=0$ has a unique solution $x_0$. In addition, it is easy to see that $g^{\prime }(5)g^{\prime }(6)<0$ so $x_0\in (5,6)$. Because $g^{\prime }(x)>0$ with $x<x_0$ and $g^{\prime }(x)<0$ with $x>x_0$, $g(x)$ reaches its maximum value at $x=x_0$. Because $k$ is a positive integer, we have $g(k)\le \max \{g(5),g(6)\}=0$. $\square$