$ĐK:x\geq 1/2$
$\sqrt{2}.\sqrt{(x+2)^2+(\sqrt{2x-1})^2}> \sqrt{2x-1}+x+2$(1)$\Leftrightarrow \sqrt{1^2+1^2}.\sqrt{(x+2)^2+(\sqrt{2x-1})^2}>\sqrt{2x-1}+x+2$
đặt $\begin{cases}\overrightarrow{u}=(1;1) \\ \overrightarrow{v}=(x+2;\sqrt{2x-1} \end{cases}$
ta luôn có $\left| {\overrightarrow{u}} \right|.\left| {\overrightarrow{v}} \right|\geq \overrightarrow{u}.\overrightarrow{v}$
suy ra $\sqrt{2}.\sqrt{(x+2)^2+(\sqrt{2x-1})^2}\geq \sqrt{2x-1}+x+2$(2)
ta thấy BPT (2) dấu = k xảy ra
nên BPT (1) luôn đúng. vậy nghiệm là x>=1/2