## Disuguaglianza

Polinomi, disuguaglianze, numeri complessi, ...
Sepp
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Iscritto il: 01 gen 1970, 01:00
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### Disuguaglianza

$\frac{1}{\sqrt{a + \frac{1}{b} + 0.64}} + \frac{1}{\sqrt{b + \frac{1}{c} + 0.64}} + \frac{1}{\sqrt{c + \frac{1}{a} + 0.64}} \geq 1.2$

con $a, b, c$ reali positivi e $abc = 1$

¬[ƒ(Gabriel)³²¹º]¼+½=¾
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Iscritto il: 22 ott 2006, 14:36
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### Re: Disuguaglianza

Ci provo...

chiamiamo $\displaystyle \left ( a + \frac{1}{b} + 0.64 \right ) = X$, $\displaystyle \left ( b + \frac{1}{c} + 0.64 \right ) = Y$, $\displaystyle \left ( c + \frac{1}{a} + 0.64 \right ) = Z$,

$\displaystyle \frac{1}{\sqrt{X}} + \frac{1}{\sqrt{Y}} + \frac{1}{\sqrt{Z}} = \frac{\sqrt{XY} + \sqrt{XZ} + \sqrt{YZ}}{\sqrt{XYZ}} = $$\displaystyle \sqrt{\frac{XY + XZ + YZ + 2X\sqrt{YZ} + 2Y\sqrt{XZ} + 2Z\sqrt{XY}}{XYZ}} Ma sviluppando \displaystyle XYZ = \left ( 1 + 0.64 + 0.64^2 \right ) \left ( a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) +$$ \displaystyle 0.64 \left ( \frac{a}{c} + \frac{c}{b} + \frac{b}{a} \right ) + 3 \cdot 0.64 + 2 + 0.64^3$

Mentre $\displaystyle XY + XZ + YX = \left ( 1 + 1.28 \right ) \left ( a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) + $$\displaystyle \left ( \frac{a}{c} + \frac{c}{b} + \frac{b}{a} \right ) + 3 \left ( 1 + 0.64^2 \right ) Mentre \displaystyle 2X\sqrt{YZ} + 2Y\sqrt{XZ} + 2Z\sqrt{XY} \geq$$ \displaystyle 2 \left ( a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} + 3 \cdot 0.64 \right )$

Essendo $\sqrt{XY}$, $\sqrt{XZ}$, $\sqrt{YZ}$ maggiori di 1

Quindi $\displaystyle \sqrt{\frac{XY + XZ + YZ + 2X\sqrt{YZ} + 2Y\sqrt{XZ} + 2Z\sqrt{XY}}{XYZ}} \geq $$\displaystyle \sqrt{\frac{4.28 \left ( a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) + \left ( \frac{a}{c} + \frac{c}{b} + \frac{b}{a} \right ) + 8.0688}{2.0496 \left ( a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ) + 0.64 \left ( \frac{a}{c} + \frac{c}{b} + \frac{b}{a} \right ) + 4.182144}} = = \displaystyle \sqrt{\frac{ 1.5625 (\mathrm{Denominatore}) + \mathrm{Roba}}{\mathrm{Denominatore}} =$$ \displaystyle \sqrt{1.5625 + \mathrm{Robaccia}} > \sqrt{1.5625} = 1.25$

Dove $\mathrm{Roba} > 0$ e $\mathrm{Robaccia} > 0$

Quindi $\displaystyle \frac{1}{\sqrt{a + \frac{1}{b} + 0.64}} + \frac{1}{\sqrt{b + \frac{1}{c} + 0.64}} + \frac{1}{\sqrt{c + \frac{1}{a} + 0.64}} > 1.25$