<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] 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.
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