I am searching for the first proof of (or counterexample to) the following conjecture. 


(The sum of squared logarithms conjecture)
For all natural numbers $n$ and positive numbers $x\_1,x\_2, \ldots , x\_n, y\_1,y\_2,\ldots, y\_n>0$ such that for all 
$k\in\{1,\ldots, n-1\}$ 
it holds


$\sum\_{i\_1<\ldots<i\_k} x\_{i\_1}\, x\_{i\_2}\ldots x\_{i\_k}\le \sum\_{i\_1<\ldots<i\_k} y\_{i\_1}\, y\_{i\_2}\ldots y\_{i\_k}$


and $x\_1\, x\_2\, x\_3 \ldots x\_n=y\_1\, y\_2 \,y\_3\ldots y\_n$


it follows 


$\sum\_{i=1}^n (\log x\_i)^2\le \sum\_{i=1}^n (\log y\_i)^2$


Replacing the assumption $x\_1\, x\_2\, x\_3 \ldots x\_n=y\_1\, y\_2\, y\_3\ldots y\_n$ by
$x\_1\, x\_2\, x\_3 \ldots x\_n\le y\_1\, y\_2\, y\_3\ldots y\_n$ easily admits counterexamples.


Proofs are known for $n\in \{1,2,3,4\}$. More information can be found at


<https://www.uni-due.de/mathematik/ag_neff/log_conjecture>


Immediately after Lev Borisov's sketch of a proof idea below, Lev Borisov, Suvrit Sra, Christian Thiel and myself agreed to work out the details and to write together a complete and self-contained paper on the sum of squared logarithm conjecture and relations to other topics which can be found at: <http://arxiv.org/abs/1508.04039>
${}{}{}$


As announced in my first post, the prize winner is Lev Borisov. 



The sum of squared logarithms conjecture
Is there anything wrong with the following argument?


First of all, by scaling all $x\_i$ and all $y\_i$ by a positive constant, we may safely assume that $\prod\_i x\_i = \prod\_i y\_i =1$.


The result would now follow from the following more general conjecture.


$\bf Conjecture:$ For ${\bf a}=(a\_1,\ldots,a\_{n-1})\in (\mathbb R\_{>0})^{n-1}$ consider
$h\_{\bf a}(z) = z^n+a\_{n-1}z^{n-1}+\cdots+a\_1z+1$. Define the function $f:(\mathbb R\_{>0})^{n-1}\to {\mathbb R}$ by
$$
f(a\_1,\ldots,a\_{n-1}) = \sum\_{z\vert h\_{\bf a}(z)=0} (\log(-z))^2
$$
where the branch of $\log$ is picked as usual on the complement of ${\mathbb R}\_{\leq 0}$ in $\mathbb C$. Then all partial derivatives $\frac{\partial f}{\partial a\_k}$ are positive.


First, a few comments. The function $f$ is well-defined because none of the roots 
of $h\_a(z)$ are positive reals. In addition, $f$ is real-valued, because roots of $h\_a(z)$ come in complex conjugate pairs for which values of $\log(-z)$ are complex conjugates.
The consequence of this conjecture is that if ${\bf a}\leq {\bf b}$ coordinate-wise,
then $f({\bf a})\leq f({\bf b})$. Applied to the case when roots of $h\_{\bf a}(z)$
and $h\_{\bf b}(z)$ are real numbers $-x\_i$ and $-y\_i$, we get the original conjecture.


Now, let me describe what I think is the proof of this new conjecture.


First of all, as standard, I can write the function $f(\bf a)$ by a contour integral. In a neighborhood of fixed ${\bf a}$ for any large enough $R$ and small enough $\epsilon>0$ there holds
$$
f({\bf a}) = \frac 1{2\pi i}\int\_{C} (\log(-z))^2 \frac {h\_{\bf a}'(z)}{h\_{\bf a}(z)}\,dz
$$
over the contour $C$ on the union of $\mathbb C$ with two copies of $\mathbb R\_{>0}$
which I will call "north" and "south" shores (so that $\log(-t)=\log t -\pi i$ on the north shore and $\log(-t)=\log t +\pi i$ on the south shore).
The contour $C$ is the union of the following four pieces $C\_\epsilon$, $C\_R$, $C\_+$, $C\_-$.


$C\_\epsilon$ is a circle of radius $\epsilon$ around $z=0$ traveled from $\epsilon$-south
to $\epsilon$-north clockwise. 


$C\_R$ is a circle of radius $R$ around $z=0$ traveled from $R$-north to $R$-south counterclockwise.


$C\_+$ is the line segment $[\epsilon,R]$-north.


$C\_-$ is the line segment $[R,\epsilon]$-south.


The derivative $\frac{\partial f({\bf a})}{\partial a\_k}$ is the integral of the 
derivative, so we get:
$$
\frac{\partial f({\bf a})}{\partial a\_k}= \frac 1{2\pi i}\int\_{C} (\log(-z))^2 \frac \partial{\partial a\_k}\frac {h\_{\bf a}'(z)}{h\_{\bf a}(z)}\,dz
$$
$$
=
\frac 1{2\pi i}\int\_{C} (\log(-z))^2 \Big(\frac {z^k}
{h\_{\bf a}(z)}
\Big)'\,dz
=
-\frac 1{2\pi i}\int\_{C} \Big((\log(-z))^2 \Big)'\frac {z^k}
{h\_{\bf a}(z)}
\,dz
$$
$$
=-\frac 1{\pi i}\int\_{C} \log(-z)\frac {z^{k-1}}
{h\_{\bf a}(z)}\,dz = -\frac 1\pi {\rm Im}\Big(\int\_{C} \log(-z)\frac {z^{k-1}}
{h\_{\bf a}(z)}\,dz\Big)
$$
We can take a limit as $R\to +\infty$ and $\epsilon\to 0$. Since $k\leq {n-1}$ the integral over $C\_R$ goes to zero (the length is $2\pi R$ and the size of the function is
$O(R^{-2}\log R)$). The integral over $C\_\epsilon$ also goes to zero, because the $k>=1$,
so the function is $O(\log \epsilon)$ and the length is $2\pi\epsilon$.


So we get 
$$
\frac{\partial f({\bf a})}{\partial a\_k}
= -\lim\_{\epsilon\to 0^+}\frac 1\pi {\rm Im}\Big( \int\_{[\epsilon,+\infty]-{\rm north}\cup [+\infty,\epsilon]-{\rm south}}
\log(-z)\frac {z^{k-1}}
{h\_{\bf a}(z)}\,dz\Big)
$$
$$
=-\lim\_{\epsilon\to 0^+}\frac 1\pi 
\int\_{\epsilon}^{+\infty}{\rm Im}\Big(
(\log(t)-\pi i)\frac {t^{k-1}}
{h\_{\bf a}(t)} -(\log(t)+\pi i)\frac {t^{k-1}}
{h\_{\bf a}(t)}\Big)\,dt
$$
$$
=2\lim\_{\epsilon \to 0^+} 
\int\_{\epsilon}^{+\infty}
\frac {t^{k-1}}{h\_{\bf a}(t)}\,dt
>0.$$
This finishes the proof of the new conjecture, and consequently of the old conjecture.


Remark:
I am guessing that there is a simpler argument for 
$$
\frac{\partial f({\bf a})}{\partial a\_k} = 2\int\_{0}^{+\infty}
\frac {t^{k-1}}{h\_{\bf a}(t)}\,dt
$$
but I am just writing the first thing that came to my mind.


