$\frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2}$ $\geq$ $\frac{3\sqrt{3}}{2}$

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hoangcoi9999

Last edited by a moderator:
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conga222222

$\eqalign{
& co: \cr
& \frac{x}{{1 - {x^2}}} \geqslant \sqrt 3 \left( {\sqrt 3 x - 1} \right) + \frac{{\sqrt 3 }}{2} = 3x - \frac{{\sqrt 3 }}{2} \cr
& \leftrightarrow \frac{x}{{1 - {x^2}}} - \frac{{\sqrt 3 }}{2} - \sqrt 3 \left( {\sqrt 3 x - 1} \right) \geqslant 0 \cr
& \leftrightarrow \frac{{2x - \sqrt 3 + \sqrt 3 {x^2}}}{{2\left( {1 - {x^2}} \right)}} - \sqrt 3 \left( {\sqrt 3 x - 1} \right) \geqslant 0 \cr
& \leftrightarrow \frac{{\left( {\sqrt 3 x - 1} \right)\left( {x + \sqrt 3 } \right)}}{{2\left( {1 - {x^2}} \right)}} - \sqrt 3 \left( {\sqrt 3 x - 1} \right) \geqslant 0 \cr
& \leftrightarrow \left( {\sqrt 3 x - 1} \right)\left( {\frac{{x + \sqrt 3 }}{{2 - 2{x^2}}} - \sqrt 3 } \right) \geqslant 0 \cr
& \leftrightarrow \left( {\sqrt 3 x - 1} \right)\left( {\frac{{2\sqrt 3 {x^2} + x - \sqrt 3 }}{{2 - 2{x^2}}}} \right) \geqslant 0 \cr
& \leftrightarrow \frac{{\left( {\sqrt 3 x - 1} \right)\left( {\sqrt 3 x - 1} \right)\left( {2x + \sqrt 3 } \right)}}{{2 - 2{x^2}}} \geqslant 0\;\left( {dung\;do\;0 < x \leqslant \frac{1}{2}} \right) \cr
& tuong\;tu... \cr
& \to VT \geqslant 3\left( {x + y + z} \right) - \frac{{3\sqrt 3 }}{2} \cr
& ma\;{\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) \geqslant 3\left( {xy + yz + zx} \right) = 3 \to x + y + z \geqslant \sqrt 3 \cr
& \to VT \geqslant 3\sqrt 3 - \frac{{3\sqrt 3 }}{2} = \frac{{3\sqrt 3 }}{2} \cr
& dau = ... \cr} $
 
H

hoangcoi9999

Ban giai thich ro hon dong dau tien , to chua hieu.........................................................................
 
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