\(\dfrac{4}{3}y=\dfrac{4x+\dfrac{8}{3}}{\sqrt{2x+1}+1}=\dfrac{4x+2+\dfrac{2}{3}}{\sqrt{2x+1}+1}\)
\(\Rightarrow\)\(\dfrac{4}{3}y=\dfrac{\left(\sqrt{2x+1}+1\right)\left(\sqrt{2x+1}-1\right)+\dfrac{2}{3}}{\sqrt{2x+1}+1}=\sqrt{2x+1}-1+\dfrac{2}{\dfrac{3}{\sqrt{2x+1}+1}}\)\(\dfrac{4}{3}y=\sqrt{2x+1}+1+\dfrac{2}{\dfrac{3}{\sqrt{2x+1}+1}}-2\)
Áp dụng BĐT Cô-si:
\(\sqrt{2x+1}+1+\dfrac{\dfrac{2}{3}}{\sqrt{2x+1}+1}>=2\sqrt{\dfrac{2}{3}}\)
\(\Rightarrow\)\(\dfrac{4}{3}y>=2\sqrt{\dfrac{2}{3}}-2\Rightarrow y>=\dfrac{2\sqrt{\dfrac{2}{3}}-2}{4}\)
Dấu = xảy ra \(\Leftrightarrow\)x=\(\dfrac{\dfrac{2}{3}-2\sqrt{\dfrac{2}{3}}}{2}\)(có thể tính sai)