$ \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 $
Disuguaglianza
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¬[ƒ(Gabriel)³²¹º]¼+½=¾
- Messaggi: 849
- Iscritto il: 22 ott 2006, 14:36
- Località: Carrara/Pisa
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 $
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 $