Inegalitate conditionata (ONM, Et.reg, Rusia-2015)

[mihai miculita]

Aratati ca:
$\left\begin{array}{c}\left(\forall\right)\,a,b,c>0\\ ab+ac+bc=1\end{array}\right\}\Rightarrow$$\sqrt{a+\dfrac{1}{a}}+\sqrt{b+\dfrac{1}{b}}+\sqrt{c+\dfrac{1}{c}}\ge2.\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right).$

[mihai miculita]

SOLUTIA din Kvant, nr.2/2015(putin mai detaliata): Avem:
$ab+ac+bc=1\Rightarrow \boxed{\sqrt{a+\dfrac{1}{a}}}=$$\sqrt{a+\dfrac{ab+ac+bc}{a}}=\sqrt{a+\dfrac{a.(b+c)+bc}{a}}=$$\sqrt{a+b+c+\dfrac{bc}{a}}=$
$=\sqrt{b+c+\left(a+\dfrac{bc}{a}\right)}\ge\sqrt{b+c+2.\sqrt{a.\dfrac{bc}{a}}}=$$\sqrt{b+c+2.\sqrt{bc}}=\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^2}=\boxed{\sqrt{b}+\sqrt{c}}\Rightarrow$
$\boxed{\Rightarrow\sqrt{a+\dfrac{1}{a}}\ge\sqrt{b}+\sqrt{c}}.\,\,(1)$
In mod analog avem, ca: $\sqrt{b+\dfrac{1}{b}}\ge\sqrt{a}+\sqrt{c};\,\,(2)$ si $\sqrt{c+\dfrac{1}{c}}\ge\sqrt{a}+\sqrt{b}.\,\,(3)$
In fine, adunand relatiile (1), (2) si (3), membru cu membru, obtinem inegalitatea din concluzia problemei.$\blacksquare$
Iata acum o varianta putin modificata a demonstratiei inegalitatii (1):
Inegalitatea conditionata: $\boxed{ab+ac+bc=1\Rightarrow\sqrt{a+\dfrac{1}{a}}\ge\sqrt{b}+\sqrt{c}},$ este echivalenta cu:
$\boxed{\sqrt{a^2+ab+ac+bc}\ge\sqrt{ab}+\sqrt{ac}}\Leftrightarrow$ $a^2+ab+ac+bc\ge\left(\sqrt{ab}+\sqrt{ac}\right)^2\Leftrightarrow$
$\Leftrightarrow a^2+ab+ac+bc\ge ab+ac+2a\sqrt{bc}\Leftrightarrow$ $a^2-2a\sqrt{bc}+bc\ge 0\Leftrightarrow \left(a-\sqrt{bc}\right)^2\ge 0.\blacksquare$