Lời giải:
Vì \(a+b+c=6\) nên BĐT cần chứng minh tương đương với:
\(\frac{ab}{2b+c+a+b+c}+\frac{bc}{2c+a+a+b+c}+\frac{ca}{2a+b+a+b+c}\leq 1(*)\)
Thật vậy, áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{ab}{2b+c+a+b+c}=\frac{ab}{(b+c)+(c+a)+2b}\leq \frac{ab}{9}\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2b}\right)\)
Hoàn toàn tương tự:
\(\frac{bc}{2c+a+a+b+c}\leq \frac{bc}{9}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{2c}\right)\)
\(\frac{ca}{2a+b+a+b+c}\leq \frac{ca}{9}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{2a}\right)\)
Cộng các BĐT vừa thu được lại ta có:
\(\text{VT}\leq \frac{1}{9}\left(\frac{ab+ac}{b+c}+\frac{ab+bc}{a+c}+\frac{bc+ca}{a+b}+\frac{a+b+c}{2}\right)\)
\(\Leftrightarrow \text{VT}\leq \frac{1}{9}\left(a+b+c+\frac{a+b+c}{2}\right)=\frac{1}{9}\left(6+\frac{6}{2}\right)=1\)
BĐT \((*)\) hoàn tất, ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c=2\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\text{VT}=\frac{ab}{6+2b+c}+\frac{bc}{6+2c+a}+\frac{ca}{6+2a+b}=\frac{ab}{a+b+c+2b+c}+\frac{bc}{a+b+c+2c+a}+\frac{ca}{a+b+c+2a+b}\)
\(=\frac{ab}{2b+(a+c)+(b+c)}+\frac{bc}{2c+(a+b)+(a+c)}+\frac{ca}{2a+(b+a)+(b+c)}\)
\(\leq \frac{ab}{9}\left(\frac{1}{2b}+\frac{1}{a+c}+\frac{1}{b+c}\right)+\frac{bc}{9}\left(\frac{1}{2c}+\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{ca}{9}\left(\frac{1}{2a}+\frac{1}{b+a}+\frac{1}{b+c}\right)\)
\(\text{VT}\leq \frac{a+b+c}{18}+\frac{ab+bc}{9(a+c)}+\frac{ab+ac}{9(b+c)}+\frac{bc+ac}{9(a+b)}\)
\(\text{VT}\leq \frac{(a+b+c)}{6}=\frac{6}{6}=1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=2$
