Lời giải:
Áp dụng BĐT AM-GM:
$3=a+b+ab\leq a+b+\frac{(a+b)^2}{4}$
$\Leftrightarrow (a+b-2)(a+b+6)\geq 0$
$\Rightarrow a+b\geq 2$
BĐT cần chứng minh tương đương với:
\(\frac{3a}{b+3}+\frac{3b}{a+3}+\frac{3ab}{a+b}\leq 3\)
\(\Leftrightarrow a-\frac{ab}{b+3}+b-\frac{ab}{a+3}+\frac{3ab}{a+b}\leq a+b+ab\)
\(\Leftrightarrow ab+\frac{ab}{a+3}+\frac{ab}{b+3}\geq \frac{3ab}{a+b}\)
\(\Leftrightarrow 1+\frac{1}{a+3}+\frac{1}{b+3}\geq \frac{3}{a+b}(*)\)
Áp dụng BĐT Cauchy-Schwarz:
\(1+\frac{1}{a+3}+\frac{1}{b+3}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{a+3}+\frac{1}{b+3}\geq \frac{36}{22+a+b}\)
\(\geq \frac{36}{11(a+b)+(a+b)}=\frac{3}{a+b}\)
BĐT $(*)$ đc chứng minh. Bài toán hoàn tất
Dấu "=" xảy r akhi $a=b=1$
