Proof of inequality by Muirhead

Question

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?

Answer 1

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.