Tìm Max

C

congchuaanhsang

$BT=\dfrac{ab}{b+1}+\dfrac{bc}{c+1}+\dfrac{ca}{a+1}$

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

$=1-(\dfrac{a^2}{ab+a}+\dfrac{b^2}{bc+b}+\dfrac{c^2}{ca+c})$

$ \le 1-\dfrac{(a+b+c)^2}{(ab+bc+ca)+(a+b+c)}$ (Cauchy-Schwarz)

\Leftrightarrow $BT \le 1-\dfrac{1}{(ab+bc+ca)+1}$ (*)

Lại có $ab+bc+ca \le \dfrac{(a+b+c)^2}{3} = \dfrac{1}{3}$

Do đó $\dfrac{1}{(ab+bc+ca)+1} \ge \dfrac{1}{\dfrac{1}{3}+1} = \dfrac{3}{4}$ (**)

Từ (*) và (**) có được $BT \le 1- \dfrac{3}{4} = \dfrac{1}{4}$

$Max=\dfrac{1}{4}$ \Leftrightarrow $a=b=c=\dfrac{1}{3}$





 
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