$f(x)=x^n+a_{n-2}x^{n-2}+...+a_1x+a_0\in \mathbb{R}[X]$. prove$\ \exists$ $i\in[1,...,n]$ so that : $$|f(x)| \ge \frac{n!}{\binom n i}$$
my attempt : i used lagrange interpolation and compared $x^n$ and $x^{n-1}$ coefficients (which is $1$ and $0$) and i wanted to use triangle inequality in the end (cus i solved something similar)
Interpolating for the coefficients of $x^{n}$ and $x^{n-1}$ respectively, the values $f\left(1\right),\dots,f\left(n\right)$ must satisfy $$n!\ =\ \left(-1\right)^{n}a_{0}\ +\ \sum_{i=1}^{n} \left(-1\right)^{n-i} \binom{n}{i} f\left(i\right)$$ and $$\binom{n}{2}\left(n-1\right)!\ =\ \left(- 1\right)^{n-1}a_{0}\ +\ \sum_{i=1}^{n-1} \left(-1\right)^{n-i-1} \binom{n-1}{i} f\left(i\right)\text{,}$$ so that $$\frac{n\left(n+1\right)}{2}\cdot \left(n-1\right)!\ =\ \sum_{i=1}^{n} \left(-1\right)^{n-i} \binom{n-1}{i-1} f\left(i\right)\text{.}$$ If, on the other hand, we have for $i=1,\dots,n$ that $$\left|f\left(i\right)\right|\ <\ \frac{n!}{\tbinom{n}{i}}\text{,}$$ then \begin{align*} \frac{n\left(n+1\right)}{2}\cdot \left(n-1\right)!\ &=\ \left|\sum_{i=1}^{n} \left(-1\right)^{n-i} \binom{n-1}{i-1} f\left(i\right)\right|\\ &\leq\ \sum_{i=1}^{n} \binom{n-1}{i-1} \left|f\left(i\right)\right|\\ &<\ \sum_{i=1}^{n} \binom{n-1}{i-1}\cdot \frac{n!}{\tbinom{n}{i}}\\ &=\ \sum_{i=1}^{n} i\cdot\left(n-1\right)!\\ &=\ \frac{n\left(n+1\right)}{2}\cdot\left(n-1\right)!\text{,} \end{align*} contradiction. $\Box$
P.s.: Given values $f\left(0\right),\dots,f\left(n\right)$, a far more efficient way to compute the minimal (i.e., at most $n$) degree interpolation of $\tilde{f}$ of $f$ (and in particular derive the expressions above) than to use Lagrange's formula is to remark that by the discrete calculus we must have $$\tilde{f}\left(x\right)\ =\ \sum_{d=0}^{n}\left(\sum_{i=0}^{d} \left(-1\right)^{d-i}\binom{d}{i} f\left(i\right)\right)\binom{x}{d}\text{,}$$ where the expression $$\sum_{i=0}^{d} \left(-1\right)^{d-i}\binom{d}{i} f\left(i\right)$$ is understood as $\left(S-1\right)^{d}$, with $$S\ \colon\ \left(\mathbb{R}\to\mathbb{R}\right)\to\left(\mathbb{R}\to\mathbb{R}\right)$$ $$S\ \colon\ \left(\varphi\ \colon\ x\mapsto \varphi\left(x\right)\right)\mapsto\left(S\varphi\ \colon\ x\mapsto\varphi\left(x+1\right)\right)$$ the "shift" operator on functions $\mathbb{R}\to\mathbb{R}$ (and $S-1$ then the discrete derivative operator on functions $\mathbb{R}\to\mathbb{R}$).