mot bai gia tri nho nhat day..

K

kakashi_hatake

$A=4(a^2+b^2+c^2)+\dfrac{1}{ab}+\dfrac{1}{bc}+ \dfrac{1}{ac} \\ a^2+b^2+c^2 \ge ab+bc+ca \\ \rightarrow A \ge 4ab+\dfrac{4}{ab}+\dfrac{4}{bc}+4bc+4ca+\dfrac{4}{ca} -3. (\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}) \\ \dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac} \le \dfrac{1}{3}. (\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2=3 \\ 4ab+\dfrac{4}{ab} \ge 2. \sqrt{4ab.\dfrac{4}{4ab}} =8
\\ \rightarrow A \ge 8.3-3.3=15$
Dấu đẳng thức xảy ra khi a=b=c=1
 
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