Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\left(2\right)\\\left(y-1\right)^2\ge0\left(3\right)\end{cases}\Leftrightarrow\hept{\begin{cases}x^2+1\ge2x\\y^2+1\ge2y\end{cases}\left(\forall x;y\inℝ\right)}}\)
\(\Rightarrow VT_{\left(1\right)}\ge\left(2x+2y+2\right)\left(2x+2y+2\right)\left(x;y\ge0\right)\)
\(\Leftrightarrow VT_{\left(1\right)}\ge4\left(x+y+1\right)^2\)(4)
Đặt \(3x+y+2=a;3y+x+b\Rightarrow a+b=4\left(x+y+1\right)\)
Lại có: \(\left(a-b\right)^2\ge0\left(\forall a;b\inℝ\right)\left(5\right)\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge ab\)
\(\Leftrightarrow\frac{16\left(x+y+1\right)^2}{4}\ge\left(3x+y+2\right)\left(3y+x+2\right)\)
\(\Leftrightarrow4\left(x+y+1\right)^2\ge\left(3x+y+2\right)\left(3y+x+2\right)=VP_{\left(1\right)}\left(6\right)\)
Từ (4) và (6) => \(VT_{\left(1\right)}\ge VP_{\left(1\right)}\)
\(\Rightarrow VT_{\left(1\right)}=VP_{\left(1\right)}\)
Dấu '=' xảy ra đồng thời ở (2), (3), (5)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=1\\3x+y+2=3y+x+2\end{cases}}\Leftrightarrow x=y=1\)
