How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$ How to prove this inequality
$$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} $$
for $a,b,c,d\gt0$?
Thanks
 A: I remember seeing this problem in some book a couple of years ago. My attention was drawn by the proof of this inequality, which apparently originates from the '70 GDR mathematical olympiad. The solution involves nothing more than AM-GM (I assume $a,b,c,d$ are nonnegative here!), but some algebraic transformations are pretty insane, so stay calm $\ddot\smile$
\begin{equation}
\begin{split}
\quad &\sqrt{\dfrac{ab + ac + ad + bc + bd + cd}{6}} = \\
&\sqrt{\dfrac{(ab+cd)/2 + (ac+bd)/2 + (ad+bc)/2}{3}} \geq \quad \text{// AM-GM applied here} \\ 
&\sqrt[6]{\dfrac{(ab+cd)(ac+bd)(ad+bc)}{8}} = \\
&\sqrt[6]{\dfrac{a^3bcd + ab^3cd + abc^3d + abcd^3}{8} + \dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{8}} = \\
&\sqrt[6]{\dfrac{\left(\dfrac{a^2 + b^2}{2} + \dfrac{b^2 + c^2}{2} + \dfrac{c^2 + d^2}{2} + \dfrac{d^2 + a^2}{2}\right)abcd + \dfrac{a^2 + c^2}{2}b^2d^2 + \dfrac{b^2 + d^2}{2}a^2c^2}{8} + }
\\
&\hspace{120pt} \overline{+ \dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{16}} \geq  \quad \text{// and here}\\
&\sqrt[6]{\dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{16} +} \\
&\hspace{120pt} \overline{+ \dfrac{2(a^2b^2cd + ab^2c^2d + abc^2d^2 + a^2bcd^2 + ab^2cd^2 + a^2bc^2d)}{16}} = \\
&\sqrt[6]{\left(\dfrac{abc + abd + acd + bcd}{4}\right)^2} = \sqrt[3]{\dfrac{abc + abd + acd + bcd}{4}} _{\square}
\end{split}
\end{equation}
Side note: if anyone knows how to improve the formatting (make the multi-line 6th root look smoother), feel free to edit this. :) I also used \overline to make the two long root expressions a bit more readable, I hope it renders properly.
A: Here is my proof for positive variables from 1979.
Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$, where $v>0$.
Hence, $a$, $b$, $c$ and $d$ are positive roots of the following equation:
$(x-a)(x-b)(x-c)(x-d)=0$ or $x^4-4ux^3+6v^2x^2-4w^3x+t^4=0$.
Hence, by Rolle the equation $(x^4-4ux^3+6v^2x^2-4w^3x+t^4)'=0$ or 
$x^3-3ux^2+3v^2x-w^3=0$ has three positive roots.
Since we can replace $x$ at $\frac{1}{x}$, we get that the equation
$w^3x^3-3v^2x^2+3ux-1=0$ has three positive roots.
Now by the Rolle's theorem again we get that the equation
$(w^3x^3-3v^2x^2+3ux-1)'=0$ or $w^3x^2-2v^2x+u=0$ has two real roots,
which says that $v^4\geq uw^3$.
By the Rolle's theorem again the equation $(x^3-3ux^2+3v^2x-w^3)'=0$
or $x^2-2ux+v^2=0$ has two positive roots, which says that $u^2\geq v^2$ or $u\geq v$.
The last inequality is also $\sum\limits_{sym}(a-b)^2\geq0$.
Id est, $v^4\geq uw^3\geq vw^3$, which gives $v\geq w$ and we are done!
A: Definitions
Suppose $p_n(x)$ is a degree $n$ polynomial with real roots, $\left\{-r_{n,k}\right\}_{k=1}^n$, where $0\lt r_{n,k}\lt r_{n,k+1}$. Define $a_k$ as follows:
$$
\begin{align}
p_n(x)
&=\prod_{k=1}^n(x+r_{n,k})\tag1\\
&=\sum_{k=0}^n\binom{n}{k}a_kx^{n-k}\tag2
\end{align}
$$
That is, $\binom{n}{k}a_k$ is the degree $k$ elementary symmetric polynomial on $\{r_{n,j}\}_{j=1}^n$.
Then, define
$$
\begin{align}
p_{n-1}(x)
&=\frac1n\,p_n'(x)\tag3\\[3pt]
&=\sum_{k=0}^{n-1}\binom{n-1}{k}a_kx^{n-1-k}\tag4\\
&=\prod_{k=1}^{n-1}(x+r_{n-1,k})\tag5
\end{align}
$$
That is, $p_{n-1}(x)$ is a degree $n-1$ polynomial. The Mean Value Theorem says that $p_{n-1}(x)$ has real roots, $\left\{-r_{n-1,k}\right\}_{k=1}^{n-1}$, where $r_{n,k}\lt r_{n-1,k}\lt r_{n,k+1}$. Furthermore, we have that $\binom{n-1}{k}a_k$ is the degree $k$ elementary symmetric polynomial on $\{r_{n-1,j}\}_{j=1}^{n-1}$.

Extension by Induction
Induction gives that for any $m\le n$,
$$
\begin{align}
p_m(x)
&=\sum_{k=0}^m\binom{m}{k}a_kx^{m-k}\tag6\\
&=\prod_{k=1}^m(x+r_{m,k})\tag7
\end{align}
$$
where $0\lt r_{m,k}\lt r_{m,k+1}$ and $\binom{m}{k}a_k$ is the degree $k$ elementary symmetric polynomial on $\{r_{m,j}\}_{j=1}^m$.
In particular, the degree $m-1$ elementary symmetric polynomial on $\{r_{m,j}\}_{j=1}^m$ is
$$
ma_{m-1}=\left(\sum_{k=1}^m\frac1{r_{m,k}}\right)\prod_{k=1}^mr_{m,k}\tag8
$$
and the degree $m$ elementary symmetric polynomial on $\{r_{m,j}\}_{j=1}^m$ is
$$
a_m=\prod_{k=1}^mr_{m,k}\tag9
$$

Inequalities for the Elementary Symmetric Functions
The AM-GM inequality gives
$$
\begin{align}
a_{m-1}
&=\frac1m\left(\sum_{k=1}^m\frac1{r_{m,k}}\right)\prod_{k=1}^mr_{m,k}\tag{10}\\
&\ge\left(\prod_{k=1}^mr_{m,k}^{-1/m}\right)\prod_{k=1}^mr_{m,k}\tag{11}\\
&=\prod_{k=1}^mr_{m,k}^{\frac{m-1}m}\tag{12}\\[6pt]
&=a_m^{\frac{m-1}m}\tag{13}
\end{align}
$$
Therefore, we have that
$$
a_{m-1}^{\ \ \frac1{m-1}}\ge a_m^{\ \frac1m}\tag{14}
$$

Answer to the Question
Apply the case $m=3$ of $(14)$ and we get $a_2^{1/2}\ge a_3^{1/3}$. Setting $n=4$, we get
$$
\left(\frac{ab+bc+ca+ad+bd+cd}6\right)^{1/2}\ge\left(\frac{abc+bcd+cda+dab}4\right)^{1/3}\tag{15}
$$
