Q:

How prove this inequality $\frac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\frac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$

let $a,b,c,d>0$, show that
$$\dfrac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\dfrac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$$
This post three -varible $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$
How prove it Four-varible ? I think use AM-GM inequality to solve it,But I can't.Thank you
I guess this follow is also true:

let $a_{1},a_{2},\cdots,a_{n}>0$,show that
  $$\dfrac{a_{1}+\sqrt{a_{1}a_{2}}+\sqrt[3]{a_{1}a_{2}a_{3}}+\sqrt[4]{a_{1}a_{2}a_{3}a_{4}}+\cdots+\sqrt[n]{a_{1}a_{2}\cdots a_{n}}}{n}\le\sqrt[n]{\dfrac{a_{1}(a_{1}+a_{2})(a_{1}+a_{2}+a_{3})\cdots(a_{1}+a_{2}+\cdots+a_{n})}{n!}}$$

Thank you

A:

\begin{align} 
a& \frac{a+b}2 \frac{a+b+c}3 \frac{a+b+c+d}4 \\
&= \frac1{4^4} \left(a+a+a+a \right) \left(a+a+b+b \right) \left(a+b+\tfrac{a+b+c}3+c \right) \left(a+b+c+d \right)  \\
&\ge \frac1{4^4} \left(a+a+a+a \right) \left(a+a+b+b \right) \left(a+b+\sqrt[3]{abc}+c \right)  \left(a+b+c+d\right) \\
&\ge \frac1{4^4} \left(a + \sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}\right)^4 \quad \text{by Holder} 
\end{align}
Not sure if one can extend the pattern to general $n$ though. For that, you may want to see http://www.jstor.org/stable/2975630

