Proof of inequality by Muirhead 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?

Two last lines in your proof are wrong.
Since $$(5,1,0)\succ(4,2,0),$$ by Muirhead $$\sum_{sym}x^5y\geq\sum_{sym}x^4y^2,$$
but you wrote a reversed inequality.
By the way, there is a proof of your inequality without expanding.
Indeed, we need to prove that
$$\sum_{cyc}\frac{yz}{2x^2+y^2+z^2}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{1}{4}-\frac{yz}{2x^2+y^2+z^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{2x^2+y^2+z^2-4yz}{2x^2+y^2+z^2}\geq0$$ or
$$\sum_{cyc}\frac{(x-y)(x+2z-y)-(z-x)(x+2y-z)}{2x^2+y^2+z^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{x+2z-y}{2x^2+y^2+z^2}-\frac{y+2z-x}{2y^2+x^2+z^2}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(2z^2-2(x+y)z+3(x^2+y^2))(2z^2+x^2+y^2)\geq0,$$ which is obvious.
