ÁD BĐT Bunhiacopxki:$(3+\frac{7}{x})^{2}=(3.1+\sqrt{7}.\frac{\sqrt{7}}{x})^{2}\leq(9+7)(1+\frac{7}{x^{2}})=16(1+\frac{7}{x^{2}})$
$\Rightarrow \sqrt{4(1+\frac{7}{x^{2}})}\geq \frac{1}{2}(3+\frac{7}{x})$
$\Rightarrow y\geq x+\frac{11}{2x}+\frac{1}{2}(3+\frac{7}{x})=\frac{3}{2}+(x+\frac{9}{x})\geq \frac{3}{2}+6=\frac{15}{2}$
Dấu''='' xra$\Leftrightarrow x=3$