Математикадан жасөспірімдер арасындағы 16-шы Балкан олимпиадасы 2012 жыл, Верия, Греция

$a$, $b$, $c$ сандары $a+b+c=1$ орындалатындай оң нақты сандар болсын. Дәлелдеңіздер: $$\dfrac {a}{b} + \dfrac {a}{c} + \dfrac {c}{b} + \dfrac {c}{a} + \dfrac {b}{c} + \dfrac {b}{a} + 6 \geq 2\sqrt{2}\left (\sqrt{\dfrac{1-a}{a}} + \sqrt{\dfrac{1-b}{b}} + \sqrt{\dfrac{1-c}{c}}\right )$$ Теңсіздік қай кезде теңдікке айналады?

Комментарий/решение

$$\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}+6\geq 2\sqrt{2}\left( \sqrt{\frac{1-a}{a}}+ \sqrt{\frac{1-b}{b}}+\sqrt{\frac{1-c}{c}}\right)$$

$$a+b+c=1\Rightarrow \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\geq 2\sqrt{2}\left( \sqrt{\frac{b+c}{a}}+ \sqrt{\frac{a+c}{b}}+\sqrt{\frac{a+b}{c}}\right)\Rightarrow$$

$$\Rightarrow \left( \frac{b+c}{a}-2\sqrt{2}\cdot \sqrt{\frac{b+c}{a}}+2 \right)+\left( \frac{a+c}{b}-2\sqrt{2}\cdot \sqrt{\frac{a+c}{b}}+2 \right)+\left( \frac{a+b}{c}-2\sqrt{2}\cdot \sqrt{\frac{a+b}{c}}+2 \right)\geq 0\Rightarrow$$

$$\Rightarrow \left(\sqrt{\frac{b+c}{a}}-\sqrt{2} \right)^2+\left(\sqrt{\frac{a+c}{b}}-\sqrt{2} \right)^2+\left(\sqrt{\frac{a+b}{c}}-\sqrt{2} \right)^2\geq 0$$

$$\left(\sqrt{\frac{b+c}{a}}-\sqrt{2} \right)^2+\left(\sqrt{\frac{a+c}{b}}-\sqrt{2} \right)^2+\left(\sqrt{\frac{a+b}{c}}-\sqrt{2} \right)^2=0 \Rightarrow$$ $$\Rightarrow \sqrt{\frac{b+c}{a}}=\sqrt{\frac{a+c}{b}}= \sqrt{\frac{a+b}{c}}=\sqrt{2}\Rightarrow \left\{ \begin{gathered} b + c = 2a\\ a+ c = 2b \\ a+ b= 2c\\ \end{gathered} \right.$$

a\b+c\b=(a+c)\b, по неравенство Коши АМ=>GM (a+c)\b+2=>2 под корню 2(a+c)\b и так следовательно со (b+c)\a+2 и (a+b)\c+2 тогда по неравенству Коши (АМ=>GM) будет решено легко.

$\sum \limits_{cyc}^{ }{\dfrac{b}{a}+\dfrac{c}{a}}+2=\sum \limits_{cyc}^{ }{\dfrac{b^2}{ab}+\dfrac{c^2}{ca}+2}\ge \sum \limits_{cyc}^{ }{\dfrac{(b+c)^2}{a(b+c)}+2}\ge \sum \limits_{cyc}^{ }{2\sqrt{2\dfrac{b+c}{a}}}=\sum \limits_{cyc}^{ }{2\sqrt{2} \Bigg(\sqrt{\dfrac{1-a}{a}}} \Bigg)$

$$(\dfrac{b}{a}+\dfrac{c}{a}+2)+(\dfrac{c}{b}+\dfrac{a}{b}+2)+(\dfrac{a}{c}+\dfrac{b}{c}+2)=(\dfrac{b^2}{ab}+\dfrac{c^2}{ca}+2)+(\dfrac{c^2}{bc}+\dfrac{a^2}{ab}+2)+(\dfrac{a^2}{ca}+\dfrac{b^2}{bc}+2)\ge$$ $$\ge \Bigg(\dfrac{(b+c)^2}{a(b+c)}+2 \Bigg)+\Bigg(\dfrac{(c+a)^2}{b(c+a)}+2 \Bigg)+\Bigg(\dfrac{(a+b)^2}{c(a+b)}+2 \Bigg)\ge 2\sqrt{2\bigg(\dfrac{b+c}{a}} \bigg)+2\sqrt{2\bigg(\dfrac{c+a}{b}} \bigg)+2\sqrt{2\bigg(\dfrac{a+b}{c}} \bigg)=2\sqrt{2} \bigg(\sqrt{\dfrac{1-a}{a}}+\sqrt{\dfrac{1-b}{b}}+\sqrt{\dfrac{1-c}{c}} \bigg) ⠀Q.E.D.$$