toán chuyên vào 10

E

eye_smile

Đặt $a=x+1;b=y+1;c=z+4$

\Rightarrow $a;b;c>0;a+b+c=6$

$Q=\dfrac{a-1}{a}+\dfrac{b-1}{b}+\dfrac{c-4}{c}=3-(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c})$

Ta có: $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{4}{c} \ge \dfrac{4}{a+b}+\dfrac{4}{c} \ge 4.\dfrac{4}{a+b+c}=\dfrac{8}{3}$

\Rightarrow $Q \le \dfrac{1}{3}$

Dấu = xảy ra \Leftrightarrow $a=b=\dfrac{3}{2};c=3$

\Leftrightarrow $x=y=\dfrac{1}{2};z=-1$
 
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