# $u_1 + \sqrt{u_1u_2} + \sqrt[3]{u_1u_2u_3} + \cdots \leq e(u_1+u_2+u_3 +\cdots)$.

$$\newcommand{\set}[1]{\{#1\}}$$ $$\newcommand{\lrp}[1]{\left(#1\right)}$$

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.

• For $n=3$ you can more more simply argue that $u_1 + \frac{1}{2}(u_1 + u_2) + \frac{1}{3}(u_1+u_2+u_3) \le (1 + \frac 1 2 + \frac 1 3)(u_1+u_2+u_3) \le e(u_1 + u_2 + u_3)$. That works for some larger $n$ as long as the partial sum of the harmonic series is less than $e$. Commented Aug 19, 2021 at 9:31
• Just for reference, the relevant keyword is the Carleman's inequality. Commented Aug 19, 2021 at 10:16

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