Giải điều kiện $x+\dfrac{1}{x}+\sqrt{x+\dfrac{1}{4}}$ ta sẽ được: $x>0$
Bình phương 2 vế:
$x^2+x+\dfrac{1}{x}+\sqrt{x+\dfrac{1}{4}}+2x\sqrt{x+\dfrac{1}{x}+\sqrt{x+\dfrac{1}{4}}}=4
\\VT \geq x^2+x+\dfrac{1}{x}+\dfrac{1}{2}+2x\sqrt{\dfrac{5}{2}}
\\=x^2+x+\dfrac{1}{x}+\dfrac{1}{2}+2x
\\=x^2+3x+\dfrac{1}{3x}+\dfrac{1}{3x}+\dfrac{1}{3x}+\dfrac{1}{2}
\\\geq 6\sqrt[6]{x^2.3x.\dfrac{1}{3x}.\dfrac{1}{3x}.\dfrac{1}{3x}}+\dfrac{1}{2}
\\=6\sqrt[6]{\dfrac{1}{9}}+\dfrac{1}{2}
>4=VP$
Do đó PTVN