\(VT\ge\left(x+y\right)+\dfrac{4}{x+y}=\left(x+y\right)+\dfrac{9}{x+y}-\dfrac{5}{x+y}\)
\(\ge2\sqrt{\left(x+y\right).\dfrac{9}{x+y}}-\dfrac{5}{3}\)\(=2.3-\dfrac{5}{3}=\dfrac{13}{3}\)
Vậy ta có đpcm. Đẳng thức xảy ra\(\Leftrightarrow x=y=\dfrac{3}{2}\)