bdt

Question

Cho $a,b,c$ là các số dương thỏa mãn: $a+b+c=1$ Tìm Max : $S = \dfrac{{ab}}{{1 + c}} + \dfrac{{ac}}{{1 + b}} + \dfrac{{bc}}{{1 + a}}$

score 4 · Answer 1

Bình chọn tăng 4 Bình chọn giảm

Bổ đề: $\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y},\forall x,y>0$
Ta có:
$S=\frac{ab}{1+c}+\frac{ac}{1+b}+\frac{bc}{1+a}$
    $=\frac{ab}{(a+c)+(b+c)}+\frac{ac}{(a+b)+(c+b)}+\frac{bc}{(a+b)+(a+c)}$
    $\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)+\frac{ac}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)+\frac{bc}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)$
    $=\frac{1}{4}\left(\frac{ac+bc}{a+b}+\frac{ab+ac}{b+c}+\frac{ab+bc}{a+c}\right)$
    $=\frac{1}{4}(a+b+c)=\frac{1}{4}$
Max$S=\frac{1}{4}\Leftrightarrow a=b=c=\frac{1}{3}$