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nttthn_97

$P^2=\dfrac{a^2b^2}{c^2}+ \dfrac{b^2c^2}{a^2}+ \dfrac{a^2c^2}{b^2}+2(\dfrac{ab}{c}.\dfrac{bc}{a}+\dfrac{ab}{c}.\dfrac{ac}{b}+\dfrac{ac}{b}.\dfrac{bc}{a})$

$=\dfrac{1}{2}[b^2(\dfrac{a^2}{c^2}+\dfrac{c^2}{a^2})+a^2(\dfrac{b^2}{c^2}+\dfrac{c^2}{b^2})+ c^2(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2})]+2(a^2+b^2+c^2)$

[TEX]\geq[/TEX]$\dfrac{1}{2}(2b^2+2a^2+2c^2)+2$[TEX]\geq[/TEX]$3$

Do a,b,c dương nên $P$[TEX]\geq[/TEX]$\sqrt{3}$
 
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