Математикадан 23-ші Балкан олимпиадасы, Агрос, Кипр, 2006 жыл

$a$, $b$ және $c$ сандары оң нақты сандар болсын. $\dfrac{1}{a(1+b)}+\dfrac{1}{b(1+c)}+\dfrac{1}{c(1+a)}\ge \dfrac{3}{1+abc}$ теңсіздігін дәлелдеңіздер.

Комментарий/решение

$\textbf{Решение №1}$ $$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\geq \frac{3}{1+abc} \Longleftrightarrow $$

$$\Longleftrightarrow (1+abc)\Bigg(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\Bigg)\geq 3\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+abc}{a(1+b)}+\frac{1+abc}{b(1+c)}+\frac{1+abc}{c(1+a)}\geq 3\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+abc}{a(1+b)}+1+\frac{1+abc}{b(1+c)}+1+\frac{1+abc}{c(1+a)}+1\geq 6\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+abc+a+ab}{a(1+b)}+\frac{1+abc+b+bc}{b(1+c)}+\frac{1+abc+c+ac}{c(1+a)}\geq 6\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+a+ab(1+c)}{a(1+b)}+\frac{1+b+bc(1+a)}{b(1+c)}+\frac{1+c+ac(1+b)}{c(1+a)}\geq 6\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+a}{a(1+b)}+\frac{ab(1+c)}{a(1+b)}+\frac{1+b}{b(1+c)}+\frac{bc(1+a)}{b(1+c)}+\frac{1+c}{c(1+a)}+\frac{ac(1+b)}{c(1+a)}\geq 6\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{(1+c)}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{(1+a)}\geq 6\Longleftrightarrow $$

$$\Longleftrightarrow \frac{1+a}{a(1+b)}+\frac{b(1+c)}{(1+b)}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{(1+c)}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{(1+a)}\geq $$

$$\geq 6\sqrt[6]{ \frac{1+a}{a(1+b)}\cdot\frac{b(1+c)}{(1+b)}\cdot\frac{1+b}{b(1+c)}\cdot\frac{c(1+a)}{(1+c)}\cdot\frac{1+c}{c(1+a)}\cdot\frac{a(1+b)}{(1+a)}}=6 $$

$$LHS\cdot abc(1+a)(1+b)(1+c)(1+abc)\ge RHS\cdot abc(1+a)(1+b)(1+c)(1+abc)$$

$$a^{3}b^{2}c^{2}+a^{2}b^{3}c^{2}+a^{2}b^{2}c^{3}+a^{2}b^{3}c+ab^{2}c^{3}+a^{3}bc^{2}-2a^{2}bc^{2}-2ab^{2}c^{2}-2a^{2}b^{2}c-2a^{2}bc-2ab^{2}c-2abc^{2}+ab^{2}+bc^{2}+ca^{2}+ab+bc+ca\ge0$$

$$ ab(b+1)(ca-1)^{2}+bc(c+1)(ab-1)^{2}+ca(a+1)(bc-1)^{2}\geq 0$$

Пусть $: \ abc=w^3 \quad a= \frac{wx}{y}, \ b = \frac{wz}{x}, \ c = \frac{wy}{z}$

\[ \sum \limits_{cyc}^{} \frac{1}{w \left ( \frac{x}{y} + \frac{wz}{y} \right )} \geq \frac{1}{w} \cdot \frac{(x+y+z)^2}{(w+1)(xy+xz+yz)}\]

\[ RHS \geq 3 \cdot \frac{1}{w(w+1)}\geq 3\cdot \frac{1}{1+w^3} \quad \square\]