Toán 12 GTNN, GTLN

T

tranduy101093

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J

jet_nguyen


 Gợi ý:
Vì $0< x,y,z <1 $ nên:
$$\left\{ \begin{array}{1} x^2 \le x \\ y^2 \le y \\ z^2 \le z \end{array}\right.$$$$ \Longleftrightarrow \left\{ \begin{array}{1} 3x^2 \le 3x\\2y^2 \le 2y\\z^2 \le z \end{array}\right.$$
Cộng 3 BĐT lại ta được $$3x^2+2y^2+z^2 \le 3x+2y+z \le4 <\dfrac{16}{3}$$
 
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