Bump... Any solution?

What a beautiful problem Label the polynomials from left to right $P_1(x),P_2(x),\dots, P_n(x)$. Then by the Cauchy–Schwarz inequality $$\sum_{k=1}^{n}|P_k(x)|\leq \sqrt{n}\cdot \sqrt{\sum_{i=1}^{n}(P_k(x))^2}$$So now it suffices to find an interval such that for all $x\in (x_i,x_{i+1})$ we have that $$Q(x)=\sum_{k=1}^{n}(P_i(x))^2<1$$Assume that this was not true for every interval. Notice that $Q(x_i)=1$ for all $i$. We wish to prove that $Q(x)-1$ has more than $2n-2$ roots counting multiplicity which would mean that $Q\equiv 1$ as $Q(x)$ has degree $2n-2$, but this is absurd taking $x\rightarrow \pm \infty$. Next to each $x_i$ on the number line write either a $+$ or a $-$ both to its left and right to signal if $Q(x)$ is increasing or decreasing as you move away from $x_i$ in either direction. Counting each $x_i$ we are at a total of $n$ already. If a certain $x_i$ has a $+$ on either side then we can increase the count by $1$ as $x_i$ must have multiplicity $2$. If an interval $(x_i,x_{i+1})$ has a $+$ and $-$ to either side then it must cross $y=1$, increasing our count by $1$. If an interval $(x_i,x_{i+1})$ has a $-$ to either side then our count increases by $2$. If $x_1$ or $x_n$ has a $-$ to its left or right, respectively, then our count increases by $1$ by considering the end behavior of $Q(x)$. Thus at each $x_i$ having two $+$ signs will increase our count by $1$ and each $-$ sign will increase our count by $1$. So in the end our count is at least $2n$, finishing the problem!

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This is a problem by Navid Safaei. His solution is very similar to #3. I don't understand the above post. Maybe it exploits something more specific, so it it's a good idea to be written in English to reach more people. The problem can be generalized a bit Claim. Let $x_1<x_2<\dots<x_n$ and $y_i\in\mathbb{R},i=1,2,\dots,n.$ Let $P(x)$ be the Lagrange interpolation polynomial $$\displaystyle P(x):= \sum_{i=1}^n y_i\ell_i(x),$$and $$\displaystyle f(x):= \sum_{i=1}^n \left| y_i\right| \left|\ell_i(x)\right|.$$Denote for brevity $x_0=-\infty, x_{n+1}=+\infty.$ Then there exists $k\in\{0,1,\dots,n\}$ such that $$\displaystyle f(x) \le \sqrt n |P(x)|\,,\, \forall x\in (x_k,x_{k+1}).$$One can see the details in this blog post.