\(3\left(x+y+1\right)^2+1\ge3xy\)
\(\Leftrightarrow3\left(x^2+y^2+1+2x+2y+2xy\right)+1\ge3xy\)
\(\Leftrightarrow3x^2+3y^2+6x+6y+6xy+4\ge3xy\)
\(\Leftrightarrow3x^2+3y^2+6x+6y+3xy+4\ge0\)
\(\Leftrightarrow3\left(x^2+\dfrac{1}{4}y^2+1+2x+xy+y\right)+\left(\dfrac{9}{4}y^2+3y+1\right)\ge0\)
\(\Leftrightarrow3\left[\left(x+1\right)^2+y\left(x+1\right)+\dfrac{1}{4}y^2\right]+\left(\dfrac{3}{2}y+1\right)^2\ge0\)
\(\Leftrightarrow3\left(x+\dfrac{y}{2}+1\right)^2+\left(\dfrac{3y}{2}+1\right)^2\ge0\left(luondung\right)\)
- Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{y}{2}+1=0\\\dfrac{3y}{2}+1=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{2}{3}\)
- Vậy bất đẳng thức đã được chứng minh.
