Desigualdades

Sendo a, b, c números reais positivos, mostre que


Dica: 

I)veja que:

[latex]2\left ( \frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c} \right ) =\frac{1}{a + b + c}\cdot \left ( \frac{2\cdot \left ( a + b + c \right )}{a + b} + \frac{2\cdot \left ( a + b + c \right )}{a + c} + \frac{2\cdot \left ( a + b + c \right )}{b + c} \right )[/latex]


[latex]2\left ( \frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c} \right ) =\frac{1}{a + b + c}\cdot \left ( 3 + \left ( \frac{a + c}{a+ b} + \frac{a + b}{a + c} \right ) + \left ( \frac{a+b}{b + c} + \frac{b + c}{a + b} \right ) + \left ( \frac{a + c}{b + c} + \frac{b + c}{a + c} \right ) \right )[/latex]


II)Pela desigualdade das medias temos que:

[latex]\left ( \frac{a + c}{a+ b} + \frac{a + b}{a + c} \right ) + \left ( \frac{a+b}{b + c} + \frac{b + c}{a + b} \right ) + \left ( \frac{a + c}{b + c} + \frac{b + c}{a + c} \right ) \geq 6\cdot \sqrt[6]{ \frac{a + c}{a+ b}\cdot \frac{a + b}{a + c}\cdot \frac{a+b}{b + c}\cdot \frac{b + c}{a + b}\cdot \frac{a + c}{b + c}\cdot \frac{b + c}{a + c}}[/latex]


[latex]\left ( \frac{a + c}{a+ b} + \frac{a + b}{a + c} \right ) + \left ( \frac{a+b}{b + c} + \frac{b + c}{a + b} \right ) + \left ( \frac{a + c}{b + c} + \frac{b + c}{a + c} \right ) \geq 6[/latex]


III)Com isso,

[latex]2\left ( \frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c} \right ) =\frac{1}{a + b + c}\cdot \left ( 3 + \left ( \frac{a + c}{a+ b} + \frac{a + b}{a + c} \right ) + \left ( \frac{a+b}{b + c} + \frac{b + c}{a + b} \right ) + \left ( \frac{a + c}{b + c} + \frac{b + c}{a + c} \right ) \right )[/latex]

[latex]2\left ( \frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c} \right ) \geq \frac{1}{a + b + c}\cdot \left ( 3 + 6 \right )[/latex]

[latex]2\left ( \frac{1}{a + b} + \frac{1}{a + c} + \frac{1}{b + c} \right ) \geq \frac{9}{a + b + c}[/latex]

Fazendo u=a+b, v=a+c, w=b+c, u, v, w positivos, é o mesmo que provar que
[latex]\frac{1}{u}+\frac{1}{v}+\frac{1}{w}\geq \frac{9}{u+v+w}[/latex], ou seja,
[latex]({u+v+w})\left(\frac{1}{u}+\frac{1}{v}+\frac{1}{w}\right)\geq 9[/latex]
mas isso é precisamente o que diz MA ≥ MH