$u_1 + \sqrt{u_1u_2} + \sqrt[3]{u_1u_2u_3} + \cdots \leq e(u_1+u_2+u_3 +\cdots)$. $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\lrp}[1]{\left(#1\right)}$

IMO Longlisted Problem 28.
Let $u_1, \ldots, u_n$ be a positive real numbers.
For each $k$ in $\set{1, \ldots, n}$ define $v_k=\sqrt[k]{u_1 \cdots  u_k}$.
Show that
$$
v_1 + \cdots + v_n \leq e(u_1 + \cdots + u_n)
$$

Attempt.
I am able to show this only for $n=1, 2, 3$.
For $n=1$ the statement is trivial.
For $n=2$, we have
$$
v_1+  v_2
= u_1 + \sqrt{u_1 u_2}
\leq u_1 + \frac{1}{2}(u_1+u_2)
\leq \lrp{1 + \frac{1}{2}} (u_1+u_2)
\leq e(u_1+u_2)
$$
For $n=3$ we have
$$
e(u_1+u_2 + u_3)
=
(u_1+u_2+u_3) + \frac{1}{2}(u_1+u_2+u_3) + \lrp{ \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots } (u_1+u_2+u_3)
$$
which gives
$$
e(u_1+u_2+u_3)
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + \lrp{ \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots }(u_1+u_2+u_3)
$$
and hence
$$
e(u_1+u_2+u_3)
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + 3\lrp{\lrp{1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots} - (1 + 1/2) } \frac{u_1+u_2+u_3}{3}
$$
Therefore
$$
e(u_1+u_2+u_3)
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + 3(e - 1.5) \frac{u_1+u_2+u_3}{3}
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + \frac{1}{3}(u_1+u_2+u_3)
$$
which using AM-GM gives the desired.

The Polya's proof in my writing.
Let $c_k=\frac{(k+1)^k}{k^{k-1}}.$
Thus,  $c_1c_2...c_k=(k+1)^k$ and by AM-GM we obtain:
$$\sum_{k=1}^n\sqrt[k]{u_1 u_2 ... u_k}=\sum_{k=1}^n\frac{\sqrt[k]{u_1c_1 u_2c_2 ... u_kc_k}}{k+1}\leq\sum_{k=1}^n\frac{u_1c_1+u_2c_2+...+u_kc_k}{k(k+1)}=$$
$$=\sum_{k=1}^n u_kc_k\sum_{i=k}^n\frac{1}{i(i+1)}=\sum_{k=1}^n u_kc_k\left(\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}+...+\frac{1}{n}-\frac{1}{n+1}\right)<$$
$$<\sum_{k=1}^nu_kc_k\cdot\frac{1}{k}=\sum_{k=1}^nu_k\frac{(k+1)^k}{k^{k-1}}\cdot\frac{1}{k}=\sum_{k=1}^nu_k\left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{n}\right)^n\sum_{k=1}^nu_k<e\sum_{k=1}^nu_k.$$
