\(\left\{{}\begin{matrix}\sqrt{2018x^2+9}\ge\sqrt{9}=3\\\sqrt{4y^2+4y+5}=\sqrt{\left(2y+1\right)^2+4}\ge\sqrt{4}=2\end{matrix}\right.\)
\(\Rightarrow VT\ge2+3=5\) (1)
\(4x^2\ge0\Rightarrow5-4x^2\le5\Rightarrow VP\le5\) (2)
Từ (1),(2) \(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra khi và chỉ khi \(VT=VP=5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2018x^2=0\\\left(2y+1\right)^2=0\\4x^2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-\frac{1}{2}\end{matrix}\right.\)