The inequality is homogeneous in $x,y,z$. Hence assume w.l.o.g. $x+y+z=1$. Thus, $xy+yz+zx \le 1/3$. Now, since $t\mapsto t^{-1/2}$ is a convex map, applying Jensen's inequality (with weights $x,y,z$), we obtain \[ \sum \frac{x}{\sqrt{y+z}} \ge \frac{1}{\sqrt{2xy+2yz+2zx}}\ge \sqrt{\frac32}. \]

jasperE3 wrote: If $x,y,z$ are positive numbers, prove that $$\frac x{\sqrt{y+z}}+\frac y{\sqrt{z+x}}+\frac z{\sqrt{x+y}}\ge\sqrt{\frac32(x+y+z)}.$$ The following inequality is also easy. Let $x$, $y$ and $z$ be positive numbers. Prove that: $$\frac x{\sqrt{y+z}}+\frac y{\sqrt{z+x}}+\frac z{\sqrt{x+y}}\ge\sqrt[4]{\frac{27(x^2+y^2+z^2)}{4}}.$$

By Hölder's inequality, we get that $$\left(\sum \frac{x}{\sqrt{y+z}}\right)^2\left(\sum x(y+z)\right) \ge \left(\sum x\right)^3,$$which is $$\sum \frac{x}{\sqrt{y+z}}\ge \sqrt{\frac{\left(\sum x\right)^3}{2 \sum xy}}.$$Let $A=\sum x^2$ and $B=\sum xy$ By AM-GM, we have \begin{align*}\sqrt{\frac{\left(\sum x\right)^3}{2 \sum xy}} &= \sqrt[4]{\frac{\left(\sum x\right)^6}{4 \left(\sum xy\right)^2}} \\&= \sqrt[4]{\frac{\left(A+B+B\right)^3}{4B^2}} \\&\ge \sqrt[4]{\frac{27AB^2}{4B^2}}\\&=\sqrt[4]{\frac{27A}{4}} \\&=\sqrt[4]{\frac{27(x^2+y^2+z^2)}{4}}.\end{align*}

jasperE3 wrote: If $x,y,z$ are positive numbers, prove that $$\frac x{\sqrt{y+z}}+\frac y{\sqrt{z+x}}+\frac z{\sqrt{x+y}}\ge\sqrt{\frac32(x+y+z)}.$$ https://artofproblemsolving.com/community/c6h71060p414066