[Toán 8] Tìm Max

E

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$6-A=(1-\dfrac{a}{1+a})+(2-\dfrac{2b}{1+b})+(3-\dfrac{3c}{1+c})=\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+b}+\dfrac{1}{1+c}+\dfrac{1}{1+c}+\dfrac{1}{1+c} \ge \dfrac{36}{6+a+2b+3c}=3$

\Rightarrow $A \le 3$

Dấu = xảy ra \Leftrightarrow $a=b=c=1$
 
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