Ta có:
$\left(3+\dfrac{7}{x}\right)^2=\left(3.1+\sqrt7.\dfrac{\sqrt7}{x}\right)^2\le(9+7)\left(1+\dfrac{7}{x^2}\right)=16\left(1+\dfrac{7}{x^2}\right)$
$\Rightarrow 2\sqrt{1+\dfrac{7}{x^2}}\ge\dfrac{1}{2}\left(3+\dfrac{7}{x}\right)$
Suy ra:
$y\ge x+\dfrac{11}{2x}+\dfrac{1}{2}\left(3+\dfrac{7}{x}\right)=\dfrac{3}{2}+x+\dfrac{9}{x}\ge\dfrac{3}{2}+2\sqrt{x.\dfrac{9}{x}}=\dfrac{15}{2}$
$\min y=\dfrac{15}{2} \Leftrightarrow x=3$