Giải thích các bước giải:
Áp dụng bđt bunhia ta có :
$(4x^2+6y^2+3z^2)(\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{3})\ge (\sqrt{4x^2.\dfrac{1}{4}}+\sqrt{6y^2.\dfrac{1}{6}}+\sqrt{3z^2.\dfrac{1}{3}})^2=(x+y+z)^2=9$
$\rightarrow(4x^2+6y^2+3z^2).\dfrac{3}{4}\ge 9$
$\rightarrow 4x^2+6y^2+3z^2\ge 12$
$\rightarrow Min 4x^2+6y^2+3z^2= 12$
Dấu = xảy ra
$\rightarrow \dfrac{4x^2}{\dfrac{1}{4}}=\dfrac{6y^2}{\dfrac{1}{6}}=\dfrac{3z^2}{\dfrac{1}{3}}$
$\rightarrow 16x^2=36y^2=9z^2$
$\rightarrow 4x=6y=3z$
$\rightarrow \dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{4}=\dfrac{x+y+z}{3+2+4}=\dfrac{1}{3}$
$\rightarrow x=1, y=\dfrac{2}{3},z=\dfrac{4}{3}$
