An inequality of symmetric polynomials using AGM inequality? Let $x_1,x_2,x_3,x_4>0$ be positive real numbers and suppose that
\begin{equation}\tag{1}
x_1x_2x_3x_4 = \frac{x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4}{6}.
\end{equation}
I want to try and show that the following inequality holds:
$$\tag{2}
1+\frac{\sqrt{x_1x_2x_3x_4}-1}{\sqrt{2}} \leq\frac{x_1+x_2+x_3+x_4}{4} \leq 1+\frac{x_1x_2x_3x_4-1}{\sqrt{2}}
$$
I believe this is true, but trying to use Lagrange multipliers isn't getting me anywhere. This conjecture was born out of investigations I was making into using the AGM inequality on symmetric polynomials with constraints. 
From the AGM inequality, we see that 
$$
\sqrt{x_1x_2x_3x_4}\leq\frac{x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4}{6}
$$
always holds for all nonnegative $x_i$. Hence if the inequality in (1) holds, then $\sqrt{x_1x_2x_3x_4}\leq x_1x_2x_3x_4$ holds. Unfortunately this doesn't help me prove (2).
I also believe that equality in either of the inequalities in (2) holds if and only if $x_1=x_2=x_3=x_4=1$. But I can't show this either. 
Can anyone help me find a proof?
 A: We'll prove a right inequality (I think the left inequality we can prove by the similar way).
Let $f(x_1,x_2,x_3,x_4)=1+\frac{x_1x_2x_3x_4-1}{\sqrt2}-\frac{x_1+x_2+x_3+x_4}{4}+\lambda\left(x_1x_2x_3x_4 -\frac{x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4}{6}\right).$
Hence, since a continuous function on the compact gets there a minimal value and if $(x_1,x_2,x_3,x_4)$ is a minimal point, we obtain:
 $$\frac{\partial f}{\partial x_1}=\frac{x_2x_3x_4}{\sqrt2}-\frac{1}{4}+\lambda\left(x_2x_3x_4 -\frac{x_2+x_3+x_4}{6}\right)=0$$ or
$$x_1x_2x_3x_4\left(\frac{1}{\sqrt2}+\lambda\right)=\frac{x_1}{4}+\frac{\lambda x_1(x_2+x_3+x_4)}{6},$$
which gives
$$\frac{x_1}{4}+\frac{\lambda x_1(x_2+x_3+x_4)}{6}=\frac{x_2}{4}+\frac{\lambda x_2(x_1+x_3+x_4)}{6}$$ or
$$(x_1-x_2)\left(\frac{3}{2}+\lambda(x_3+x_4)\right)=0.$$
Similarly we obtain in the minimal point:
$$(x_1-x_3)\left(\frac{3}{2}+\lambda(x_2+x_4)\right)=0,$$
$$(x_1-x_4)\left(\frac{3}{2}+\lambda(x_2+x_3)\right)=0,$$
$$(x_2-x_3)\left(\frac{3}{2}+\lambda(x_1+x_4)\right)=0,$$
$$(x_2-x_4)\left(\frac{3}{2}+\lambda(x_1+x_3)\right)=0,$$
$$(x_3-x_4)\left(\frac{3}{2}+\lambda(x_1+x_2)\right)=0.$$
Now, if $x_1\neq x_2$, $x_1\neq x_3$ and $x_1\neq x_4$ so we get $x_2=x_3=x_4$.
If $x_1=x_2$, but $x_1\neq x_3$ and $x_1\neq x_4$ so $x_3=x_4$.
Thus, it remains to prove our inequality in the following cases.


*

*$x_1=x_2=x_3=x_4$;

*$x_1=a$ and $x_2=x_3=x_4=b$;

*$x_1=x_2=a$ and $x_3=x_4=b$.
The rest for you.
