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

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.

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    $\begingroup$ 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 $e$. $\endgroup$
    – Martin R
    Commented Aug 19, 2021 at 9:31
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    $\begingroup$ Just for reference, the relevant keyword is the Carleman's inequality. $\endgroup$ Commented Aug 19, 2021 at 10:16

1 Answer 1

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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: $$\sum_{k=1}^n\sqrt[k]{u_1 u_2 ... u_k}=\sum_{k=1}^n\frac{\sqrt[k]{u_1c_1 u_2c_2 ... u_kc_k}}{k+1}\leq\sum_{k=1}^n\frac{u_1c_1+u_2c_2+...+u_kc_k}{k(k+1)}=$$ $$=\sum_{k=1}^n u_kc_k\sum_{i=k}^n\frac{1}{i(i+1)}=\sum_{k=1}^n u_kc_k\left(\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}+...+\frac{1}{n}-\frac{1}{n+1}\right)<$$ $$<\sum_{k=1}^nu_kc_k\cdot\frac{1}{k}=\sum_{k=1}^nu_k\frac{(k+1)^k}{k^{k-1}}\cdot\frac{1}{k}=\sum_{k=1}^nu_k\left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{n}\right)^n\sum_{k=1}^nu_k<e\sum_{k=1}^nu_k.$$

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    $\begingroup$ If this is a known proof technique (“Pólya's proof”) then you can perhaps add a reference to the source. $\endgroup$
    – Martin R
    Commented Aug 19, 2021 at 9:43
  • $\begingroup$ Page 249-250, Hardy G.H., Littlewood J.E., Polya G., "Inequalities," 2nd Ed. 1952. $\endgroup$
    – River Li
    Commented Aug 19, 2021 at 14:09
  • $\begingroup$ @River Li Thank you! I took it from the lecture, when studied at university. $\endgroup$ Commented Aug 19, 2021 at 14:20
  • $\begingroup$ @MichaelRozenberg I saw the use of this technique several times in MSE etc. $\endgroup$
    – River Li
    Commented Aug 19, 2021 at 14:23
  • $\begingroup$ @River Li Yes, it's a Bacteria's method. The following inequality I proved also by this method: math.stackexchange.com/questions/4224749 $\endgroup$ Commented Aug 19, 2021 at 14:26

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