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huynhbachkhoa23

Đăt $t=x+y+z \rightarrow \dfrac{t^2-3}{2}=xy+yz+zx$
$$x+y+z=\sqrt{x^2+y^2+z^2+2(xy+yz+zx)} \ge \sqrt{3}$$
$$x+y+z\le \sqrt{3(x^2+y^2+z^2)}=3$$
Khảo sát hàm số.
 
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