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huynhbachkhoa23

$(x+y+z)^2=3(xy+yz+zx)+\dfrac{1}{2}\left [(x-y)^2+(y-z)^2+(z-x)^2\right] \ge 3(xy+yz+zx)$
Áp dụng vào: $18=3(x+y+z+xy+yz+zx) \le 3(x+y+z)+(x+y+z)^2$

$\leftrightarrow (x+y+z+6)(x+y+z-3) \ge 0$

$\rightarrow |x+y+z| \ge 3$

$x^2+y^2+z^2 \overset{Cauchy-Schwarz}{\ge} \dfrac{(x+y+z)^2}{3} \ge 3$
 
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