Азия-тынық мұхит математикалық олимпиадасы, 2018 жыл

$f(x)$ және $g(x)$ функциялары келесідей берілген: $f(x) = \dfrac{1}{x} + \dfrac{1}{x-2} + \dfrac{1}{x-4} + \cdots + \dfrac{1}{x-2018}$ және $g(x) = \dfrac{1}{x-1} + \dfrac{1}{x-3} + \dfrac{1}{x-5} + \cdots + \dfrac{1}{x-2017}.$ Барлық бүтін емес $0 < x < 2018$ шартын қанағаттандыратын $x$ саны үшін $|f(x) - g(x)| > 2$ теңсіздігін дәлелдеңіз. ( Australia )

Комментарий/решение

$$ |f(x)-g(x)|=\left| \frac{1}{x}+ \frac{1}{x-2}+ \frac{1}{x-4}+...+ \frac{1}{x-2018}-\left( \frac{1}{x-1}+ \frac{1}{x-3}+ \frac{1}{x-5}+...+ \frac{1}{x-2017}\right)\right|=$$

$$=\left| \frac{1}{x}+\left( \frac{1}{x-2018}- \frac{1}{x-1} \right)+\left( \frac{1}{x-2016}- \frac{1}{x-3} \right)+...+\left( \frac{1}{x-2}- \frac{1}{x-2017} \right)\right|=$$

$$=\left| \frac{1}{x}+\frac{2017}{x^2-2019x+2018}+\frac{2013}{x^2-2019x+6048}+...-\frac{2015}{x^2-2019+4034}\right|=$$

$$=\left| \frac{1}{x}+\frac{2017}{x^2-2019x+2018}+\frac{2013}{x^2-2019x+6048}+...+\frac{1}{x^2-2019x+1010\cdot 1009}-\frac{3}{x^2-2019x+1008\cdot 1011}-...-\frac{2015}{x^2-2019x+4034}\right|=$$

$$=\left| \frac{1}{x}+\underbrace{\frac{2017}{x^2-2019x+2018}+\frac{2013}{x^2-2019x+6048}+...+\frac{1}{x^2-2019x+1010\cdot 1009}}_A-\left(\underbrace{\frac{3}{x^2-2019x+1008\cdot 1011}+...+\frac{2015}{x^2-2019x+4034}}_B\right)\right|$$

$$ x^2-2019x+2018<x^2-2019x+6048<...<x^2-2019x+1010\cdot 1009\Rightarrow A>\frac{1+5+9+...+2013+2017}{x^2-2019x+1010\cdot1009}=\frac{505\cdot1009}{x^2-2019x+1010\cdot1009}$$

$$ x^2-2019x+1008\cdot1011>...>x^2-2019x+4034 \Rightarrow B<\frac{3+7+...+2015}{x^2-2019x+4034}=\frac{1009\cdot504}{x^2-2019x+4034}\Rightarrow -B>-\frac{1009\cdot504}{x^2-2019x+4034}$$

$$|f(x)-g(x)|=\left| \frac{1}{x}+A-B\right|>\left| \frac{1}{x}+\frac{505\cdot1009}{x^2-2019x+1010\cdot1009}-\frac{1009\cdot504}{x^2-2019x+4034}\right|=h(x)>\min_{x\in(0,2018)}h(x)$$

$$\min_{x\in(0,2018)}h(x)=?$$

$$\frac{505\cdot1009}{x^2-2019x+1010\cdot1009}-\frac{1009\cdot504}{x^2-2019x+4034}=Q \Rightarrow h(x)=\left|\frac{1}{x}+Q(x)\right|$$

$$Q(x)=\frac{505\cdot1009}{x^2-2019x+1010\cdot1009}-\frac{1009\cdot504}{x^2-2019x+4034}$$

$$ x^2+2019x=z\Rightarrow Q=\frac{505\cdot1009}{z+1010\cdot1009}-\frac{1009\cdot504}{z+4034}$$

$$ 505=k\Rightarrow Q=\frac{k(2k-1)}{z+2k(2k-1)}-\frac{(2k-1)(k-1)}{z+2(4k-3)}=\frac{k(2k-1)(z+2(4k-3))-(2k-1)(k-1)(z+2k(2k-1))}{(z+2k(2k-1))(z+2(4k-3))}=$$

$$=\frac{(2k-1)\left( kz+2k(4k-3)-(k-1)z-2k(2k-1)(k-1)\right)}{(z+2k(2k-1))(z+2(4k-3))}=\frac{(2k-1)\left( z-4k^3+14k^2-8k \right)}{z^2+z(6k-6+4k^2)+4k(8k^2-10k+3)}$$

$$Qz^2+Qz(6k-6+4k^2)+4kQ(8k^2-10k+3)-(2k-1)z+(2k-1)(4k^3-14k^2+8k)=0$$

$$ Qz^2+z\left(Q(6k-6+4k^2)-2k+1\right)+\left(4kQ(8k^2-10k+3)+(2k-1)(4k^3-14k^2+8k)\right)=0$$

$$D^2=\left(Q(6k-6+4k^2)-2k+1\right)^2-4Q\left(4kQ(8k^2-10k+3)+(2k-1)(4k^3-14k^2+8k)\right)\geq0$$

$$ Q^2\left((6k-6+4k^2)^2-16k(8k^2-10k+3)\right)-2Q(2k-1)\left(2(4k^3-14k^2+8k)+(6k-6+4k^2)\right)+(2k-1)^2\geq0$$

$$ k=505 \Rightarrow Q\geq \frac{1009\left(1009+12\cdot \sqrt{7070}\right)}{1015056}>2$$

$$|f(x)-g(x)|=h(x)=\left|\frac{1}{x}+Q(x)\right|>\left|\frac{1}{x}+2\right|>2$$