Homogenizing the inequality, let $x+y+z=1$, then since $f(x)=\frac{1}{\sqrt x}$ has $f''(x)=\frac{3}{4}x^{-\frac{5}{2}}>0$, $f$ is convex, by Weighted Jensen's Inequality, we have \[\frac {x}{\sqrt{yz+4xy+4xz}}+\frac {y}{\sqrt{zx+4yz+4yx}}+\frac {z}{\sqrt{xy+4zx+4zy}}\geq \frac{1}{\sqrt{4(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)+3xyz}}.\]It is suffice to show that \[(x+y+z)^3\ge 4(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)+3xyz\]wihich is equivalent to showing \[x^3+y^3+z^3+3xyz \ge x^2y+x^2z+y^2x+y^2z+z^2x+z^2y\]and this is just Shur's 3rd degree Inequality. $\ \blacksquare$

Call the given object as $S$; our goal is to show $S^2\ge 1$. Notice, by Holder's inequality, that \[ S^2\left(\sum x\bigl(yz+4xy+4xz\bigr)\right)\ge (x+y+z)^3. \]Since \[ (x+y+z)^3 = \sum x^3 + 6xyz + 3\sum_{{\rm sym}} x^2y \quad\text{and}\quad \sum x\bigl(yz+4xy+4xz\bigr) = 3xyz + 4\sum_{{\rm sym}} x^2y \]it suffices to prove \[ x^3+y^3+z^3+3xyz\ge \sum_{{\rm sym}} x^2y, \]which follows directly, as for all $r\ge 0$, Schur's inequality asserts that \[ \sum x^r(x-y)(x-z)\ge 0. \]

Notice that $\sum_{cyc}\frac{x}{\sqrt{yz+4xy+4zx}}=\sum_{cyc}\frac{x^{3/2}}{\sqrt{xyz+4x^2y+4zx^2}}=\sum_{cyc}\frac{x^{3/2}}{(xyz+4x^2+4y^2)^{1/2}}\overset{\text{Generalized T2'S}}{\ge}\frac{\left(\sum_{cyc}a\right)^{3/2}}{\left(4\sum_{sym}x^2y+3xyz\right)^{1/2}}$ Therefore we can rewrite the inequality as $\sqrt{\frac{\left(\sum_{cyc}x\right)^3}{4\sum_{sym}x^2y+3xyz}}\ge1\Longleftrightarrow\left(\sum_{cyc}x\right)^3\ge4\sum_{sym}x^2y+3xyz$ However notice that $\left(\sum_{cyc}x\right)^3=\sum_{cyc}x^3+3\sum_{cyc}x\sum_{cyc}xy-3xyz=\sum_{cyc}x^3+3\sum_{sym}x^2y+6xyz$ Thus by plugging this in we obtain $\sum_{cyc}x^3+3\sum_{sym}x^2y+6xyz\ge4\sum_{sym}x^2y+3xyz\Longrightarrow\sum_{cyc}x^3+3xyz\ge\sum_{sym}x^2y$ Which is just $\text{Schur}$ when $r=1$ $\blacksquare$.

dendimon18 wrote: Suppose $x,y,z\in \mathbb R^+$. Prove that \[\frac {x}{\sqrt{yz+4xy+4xz}}+\frac {y}{\sqrt{zx+4yz+4yx}}+\frac {z}{\sqrt{xy+4zx+4zy}}\geq 1\]. https://artofproblemsolving.com/community/c6h483086