[toán 9] tìm min

E

eye_smile

Ta có:

$18 \ge x^2+y^2+z^2+x+y+z \ge x+y+z+\dfrac{(x+y+z)^2}{3}$

Đặt $x+y+z=A$

\Rightarrow $A+\dfrac{A^2}{3} \le 18$

Giải ra đc $A \le 6$

Hay $x+y+z \le 6$

Ta có: $B=\dfrac{1}{x+y+1}+\dfrac{1}{y+z+1}+\dfrac{1}{z+x+1} \ge \dfrac{9}{2(x+y+z)+3} \ge \dfrac{9}{2.6+3}=\dfrac{3}{5}$

Dấu "=" xảy ra \Leftrightarrow $x=y=z=2$
 
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