cho a,b,c dương, a+b+c=3. CMR:
$\sum \frac{a(a-2b+c)}{ab+1}\geq 0$ (*)
p/s:giải vầy đúng không m.n?
+ ta có: (*) tương đương
$\sum \frac{a^2}{ab+1}+\sum \frac{ac}{ab+1}+2\sum \frac{1}{ab+1}\geq 6$
+ theo gt, có :a+b+c=3 => ab+bc+ac<=3
+ mặt khác:
$\sum \frac{a^2}{ab+1}\geq \frac{(a+b+c)^2}{ab+bc+ac+3}=\frac{3}{2}$
$2\sum \frac{1}{ab+1}\geq \frac{18}{ab+bc+ac+3}=3$
=> cần CM: $\sum \frac{ac}{ab+1}\geq \frac{3}{2}$
+ ta có:
$\frac{ac}{ab+1}+\frac{ab+1}{4a}+\frac{c^2}{2}\geq \frac{3c}{2}$
$\frac{bc}{ac+1}+\frac{ac+1}{4c}+\frac{b^2}{2}\geq \frac{3b}{2}$
$\frac{ab}{bc+1}+\frac{bc+1}{4b}+\frac{a^2}{2}\geq \frac{3a}{2}$
cộng vế theo vế có:
$\sum \frac{ac}{ab+1}+\frac{1}{4}\sum \frac{1}{a}+\sum \frac{a^2}{2}+\sum \frac{a}{4}\geq \sum \frac{3a}{2}=>\geq \sum \frac{ac}{ab+1}+\frac{3}{4}+\frac{3}{2}\geq \frac{15}{4}=>\sum \frac{ac}{ab+1}\geq \frac{3}{2},(dpcm))$