We have to prove: $$\frac{\sqrt{pq}}{p+q+2r}+\frac{\sqrt{pr}}{p+r+2q}+\frac{\sqrt{pr}}{p+r+2q}\leq\frac{3}{4}$$
By multiplying it all out we get the following equivalent: \begin{align*} 4\sum_{cyc}{\sqrt{pq}(p+r+2q)(q+r+2p)}\leq \\ 3(p+r+2q)(p+q+2r)(r+q+2p) \end{align*} Let us put: \begin{align*} x=\sqrt{p}\\ y=\sqrt{q}\\ z=\sqrt{r}\\ \end{align*} Now, with help of Wolfram Alpha we rewrite the equation in Muirhead-Notation: \begin{align*} 4(2[5,1,0]+\frac{1}{2}[4,1,1]+3[3,2,1]+\frac{5}{2}[3,3,0])\\ =8[5,1,0]+2[4,1,1]+12[3,2,1]+5[3,3,0]\\ \leq 3[6,0,0]+21[4,2,0]+12[2,2,2] \end{align*} We have: \begin{align*} [6,0,0]+[2,2,2]\geq 2[4,1,1]\\ 2([6,0,0]+[4,2,0])\geq 4[5,1,0]\\ 6([4,2,0]+[2,2,2])\geq 12[3,2,1]\\ 10 [4,2,0]\geq 10[3,3,0]\\ \end{align*} Where we use $\frac{[p]+[q]}{2}\geq [\frac{p+q}{2}]$ multiple times. Now we still have to prove: $4[5,1,0]\leq 3[4,2,0]+6[2,2,2]$ Now the LHS of the inequality is homogenous in $x,y,z$ so we scale them so that $\max (x,y,z)\leq \frac{3}{4}$. Now: $[5,1,0]\leq [5,2,0]\leq \frac{3}{4}[4,2,0]\leq \frac{3}{4}[4,2,0]+6[2,2,2]$.
The entirety of this proof seems rather fishy to me. Is it correct? If no, where did I go wrong?