Prove that $\frac{1}{\sqrt x}\ge\frac{2}{x+1}$ for all $x\gt0$
I started by letting $\sqrt x=n$, giving me $\frac{1}{n}\ge\frac{2}{n^2+1}$, then taking the inverse and rearranging to get the quadratic $n^2-2n+1\ge0$, which when I solve it, gives me $n\ge1$ and therefore $x\ge1$. This proves that the inequality holds for all $x\ge1$, but doesn't account for when $0\lt x\lt1$? How can I account for that?
Thanks very much in advance