Echivalenta

[mihai miculita]

Daca $a;b;c>0,$ aratati ca: $\dfrac{a^2+b^2}{c}+\dfrac{a^2+c^2}{b}+\dfrac{b^2+c^2}{a}$$=2(a+b+c)$$\Leftrightarrow a=b=c.$

[dangerous storm]

Trebuie sa aplicam inegalitatea aranjamentelor de 2 ori si obtinem rezultatul dorit.

[mihai miculita]

Altfel, folosind inegalitatea MA-MG, avem:
$a+b+c=\dfrac{a^2+b^2}{2c}+\dfrac{a^2+c^2}{2b}+\dfrac{b^2+c^2}{2a}\ge$$\dfrac{ab}{c}+\dfrac{ac}{b}+\dfrac{bc}{a}=abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\Rightarrow$
$\Rightarrow \left a+b+c\ge abc.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\right|.\dfrac{1}{abc}\Leftrightarrow$$\boxed{\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\ge \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}.\,\,(1)$
Pe de alta parte, are loc inegalitatea (cu egalitate, doar in cazul $a=b=c\,$):
$\boxed{\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\le \dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2};(\forall)a,b,c\in\mathbb{R}^*.}\,\,(2)$
In fine, din (1) si (2), urmeaza ca:
$\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}$$\Leftrightarrow \boxed{a=b=c}.$

[mihaith]

Putem aplica si inegalitatea CBS in forma Titu pentru cele 6 numere si vom obtine cazul de egalitate: a=b=c.

[Virgil Nicula]

Vedeti PP10 de aici.