<s>[INST] 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
 [/INST] 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!
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