1-ші халықаралық Жәутіков олимпиадасы, үлкен лига, 2005 жыл

Кез келген оң нақты $a$, $b$, $c$ және $d$ сандары үшін $\dfrac{c}{a+2b}+\dfrac{d}{b+2c}+\dfrac{a}{c+2d}+\dfrac{b}{d+2a}\ge \dfrac{4}{3}$. теңсіздігін дәлелдеңдер.

Комментарий/решение

$$\left\{ \begin{gathered} a + 2b = x\\ b + 2c = y \\ c + 2d = z \\ d+2a=m \\ \end{gathered} \right.\Rightarrow$$

$$\Rightarrow a=\frac{2y-4z+8m-x}{15}, b=\frac{2z-4m+8x-y}{15},$$ $$c=\frac{2x-4y+8z-m}{15},d=\frac{2m-4x+8y-z}{15}\Rightarrow$$

$$\frac{c}{a+2b}+\frac{d}{b+2c}+\frac{a}{c+2d}+\frac{b}{d+2a}=$$ $$=\frac{2y-4z+8m-x}{15x}+\frac{2z-4m+8x-y}{15y}+$$ $$+\frac{2x-4y+8z-m}{15z}+\frac{2m-4x+8y-z}{15m}=$$

$$=\frac{2}{15}\cdot \left(\frac{y}{x}+\frac{x}{z}+\frac{z}{y}\right) +\frac{2}{15}-\frac{4}{15}\left(\frac{z}{x}+\frac{x}{m}+\frac{m}{y}+\frac{y}{z}\right)+$$

$$+\frac{8}{15}\cdot \left(\frac{m}{x}+\frac{x}{y}+\frac{y}{m}\right)+\frac{8}{15}-\frac{4}{15}\geq$$

$$\geq \frac{2}{5}+\frac{2}{15}-\frac{16}{15}+\frac{8}{5}+\frac{8}{15}-\frac{4}{15}=\frac{4}{3}$$

Формула Коши. Тогда левая часть больше и равно $(a+b+c+d)^2$

$3(a+b+c+d)^2>_=8(ab+ac+ad+bc+cd+bd)$

А это у нас $A.M>_=G.M$

$$(\sum^{⠀}_{cyc}{\dfrac{c}{a+2b}})\cdot (\sum^{⠀}_{cyc}{c(a+2b)}) \ge (a+b+c+d)^2; \sum^{⠀}_{cyc}{\dfrac{a^2+b^2}{3}}+\sum^{⠀}_{cyc}{2ab} \ge \sum^{⠀}_{cyc}{\dfrac{8ab}{3}}$$