Răspuns:
Explicație pas cu pas:
La 86 se aplica inegalitatea dintre media aritmetica si cea geometrica :
[tex]\dfrac{a+b}{2}\geq \sqrt{ab} \Leftrightarrow a+b\geq 2\sqrt{ab} \Leftrightarrow (a+b)\sqrt{c}\geq 2\sqrt{abc}\\\texttt{In mod analog obtinem : }\\(b+c)\sqrt a\geq 2\sqrt{abc} \texttt{ si } (a+c)\sqrt{b}\geq 2\sqrt{abc}\\\texttt{Prin adunarea celor trei inegalitati se ajunge la : }\\(a+b)\sqrt{c}+(b+c)\sqrt{a}+(a+c)\sqrt{b}\geq 2\sqrt{abc}+2\sqrt{abc}+2\sqrt{abc}=6\sqrt{abc}[/tex]
La 87 mai intai vom face un artificiu de calcul :
[tex](a+b+c)(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9\sqrt{abc}|:\sqrt{abc}\\\left(\dfrac{a+b+c}{\sqrt{abc}}\right)(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9\\\left(\sqrt{\dfrac{a}{bc}}+\sqrt{\dfrac{b}{ac}}+\sqrt{\dfrac{c}{ab}\right)(\sqrt a+\sqrt b+\sqrt c)\geq 9\\\texttt{Aplicam inegalitatea mediilor (media aritmetica si }\\\texttt{media geometrica) pentru fiecare paranteza.}\\\sqrt{\dfrac{a}{bc}}+\sqrt{\dfrac{b}{ac}}+\sqrt{\dfrac{c}{ab}}\geq 3\sqrt[3]{\sqrt{\dfrac{abc}{a^2b^2c^2}}}=3\sqrt[3]{\sqrt{\dfrac{1}{abc}}}[/tex]
[tex]\texttt{Pentru cea de-a doua : }\\\sqrt{a}+\sqrt{b}+\sqrt{c}\geq 3\sqrt[3]{\sqrt{abc}}\\\texttt{Prin inmultirea celor doua inegalitati se obtine }\\\texttt{concluzia problemei.}\\\left(\sqrt{\dfrac{a}{bc}}+\sqrt{\dfrac{b}{ac}}+\sqrt{\dfrac{c}{ab}}\right)(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 9\sqrt[3]{\sqrt{\dfrac{1}{abc}}}\cdot\sqrt[3]{\sqrt{abc}}=\\= 9,~Q.E.D.[/tex]
