[toán 9 ] c/m:

E

eye_smile

GT \Leftrightarrow $\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=3$

Đặt $\dfrac{1}{a}=x;\dfrac{1}{b}=y;\dfrac{1}{c}=z$ \Rightarrow $xy+yz+zx=3$


\Rightarrow $BT=x^3+y^3+z^3 \ge \dfrac{(x^2+y^2+z^2)^2}{x+y+z} \ge \dfrac{(x^2+y^2+z^2)^2}{\sqrt{3}.\sqrt{x^2+y^2+z^2}} \ge 3$

Dấu = xảy ra \Leftrightarrow $x=y=z=1$

\Leftrightarrow $a=b=c=1$
 
L

lp_qt

GT \Leftrightarrow $\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=3$

$\dfrac{1}{a^3}+\dfrac{1}{b^3}+1 \ge 3.\dfrac{1}{ab}$

tuong tu, ta co:

$2.(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3})+3 \ge 3.(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac})$

\Rightarrow $\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3} \ge 3$
 
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