\(3=a+b+ab\le a+b+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\left(a+b\right)^2+4\left(a+b\right)-12\ge0\)
\(\Leftrightarrow\left(a+b-2\right)\left(a+b+6\right)\ge0\)
\(\Leftrightarrow a+b-2\ge0\Rightarrow a+b\ge2\)
Ta có:
BĐT\(\Leftrightarrow\frac{3a^2+3a+3b^2+3b}{\left(b+1\right)\left(a+1\right)}+\frac{ab}{a+b}\le a^2+b^2+\frac{3}{2}\)
\(\Leftrightarrow\frac{3a^2+3b^2+3a+3b}{4}+\frac{ab}{a+b}\le a^2+b^2+\frac{3}{2}\)
\(\Leftrightarrow3a+3b+\frac{4ab}{a+b}\le a^2+b^2+6\)
\(\Leftrightarrow3a+3b+\frac{4ab}{a+b}\le a^2+b^2+2\left(ab+a+b\right)\)
\(\Leftrightarrow a+b+\frac{4ab}{a+b}\le\left(a+b\right)^2\)
Ta có:
\(VT=a+b+\frac{4ab}{a+b}\le a+b+\frac{\left(a+b\right)^2}{a+b}=2\left(a+b\right)\le\left(a+b\right)\left(a+b\right)=\left(a+b\right)^2\)
Dấu "=" xảy ra khi \(a=b=1\)
