Desigualdades

por jaques104 Qua 08 Jul 2015, 23:05

Sejam a, b e c reais positivos tais que abc=1. Mostre que:

Desigualdades E4f49d18278fa3e7933dde0b4c7a0f9c

por **fantecele** Qui 20 Dez 2018, 02:06

Uma maneira é usando o "lema de Titu" (Titu's lemma) depois aplicando MA ≥ MG, veja:

$\small \\\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}=\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\\\\\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{b}}+\frac{a}{\sqrt{c}}+\frac{b}{\sqrt{c}}\geq \frac{(2.(\sqrt{a}+\sqrt{b}+\sqrt{c}))^2}{2.(\sqrt{a}+\sqrt{b}+\sqrt{c})}=2.(\sqrt{a}+\sqrt{b}+\sqrt{c})\\\\\\2.(\sqrt{a}+\sqrt{b}+\sqrt{c})=\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3\\\\\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Uma outra maneira é aplicando apenas MA≥ MG.

$\small \\\frac{b+c}{\sqrt{a}}\geq 2\sqrt{\frac{bc}{a}}\\\frac{c+a}{\sqrt{b}}\geq 2\sqrt{\frac{ac}{b}}\\\frac{a+b}{\sqrt{c}}\geq 2\sqrt{\frac{ab}{c}}\\\\\\\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq 2.\left ( \sqrt{\frac{bc}{a}}+\sqrt{\frac{ac}{b}}+\sqrt{\frac{ab}{c}} \right )$
$\small \\2.\left ( \sqrt{\frac{bc}{a}}+\sqrt{\frac{ac}{b}}+\sqrt{\frac{ab}{c}} \right )=\left ( \sqrt{\frac{bc}{a}}+\sqrt{\frac{ac}{b}} \right )+\left ( \sqrt{\frac{bc}{a}}+\sqrt{\frac{ab}{c}} \right )+\left ( \sqrt{\frac{ac}{b}}+\sqrt{\frac{ab}{c}} \right )\\\\\left ( \sqrt{\frac{bc}{a}}+\sqrt{\frac{ac}{b}} \right )+\left ( \sqrt{\frac{bc}{a}}+\sqrt{\frac{ab}{c}} \right )+\left ( \sqrt{\frac{ac}{b}}+\sqrt{\frac{ab}{c}} \right )\geq 2.(\sqrt{a}+\sqrt{b}+\sqrt{c})\\\\2.(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3\\\\\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Desigualdades

Desigualdades

Re: Desigualdades

Um fórum livre e democrático.