[toán 9] c/m

H

hien_vuthithanh

đặt $\dfrac{1}{a}=x,\dfrac{1}{b}=y,\dfrac{1}{c}=z$
dựa vào dk\Rightarrow xy+yz+zx=3
$x^3+y^3+z^3$=$\dfrac{x^4}{x}+\dfrac{y^4}{y}+ \dfrac{z^4}{z}$
\geq $\dfrac{(x^2+y^2+z^2)^2}{x+y+z}$ \geq $\dfrac{(x^2+y^2+z^2)^2}{\sqrt{3(x^2+y^2+z^2)}}$ \geq $\dfrac{\sqrt{(x^2+y^2+z^2)^3}}{\sqrt{3}}$ \geq $\dfrac{\sqrt{(xy+yz+zx)^3}­}{\sqrt{3}}$ = $\dfrac{\sqrt{3}^3}{\sqrt{3}}$=3
\Rightarrow dpcm
 
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