Let $X$ be a Banach space with $f_1,..,f_n\in X^*$ and $c_1,...,c_n\in\mathbb{R}$ then the following are equivalent:
1. $\exists x_0\in X: f_i(x_0)=c_i,\forall i\in\{1,..,n\}$
2. $\exists M\geq 0: |\sum_{i=1}^na_ic_i|\leq M\|\sum_{i=1}^na_if_i\|$ for every choice of $a_i\in\mathbb{R}$
The direction $1\implies 2$ is immediate by setting $f=\sum_{i=1}^na_if_i$ and noticing that $\|f\|\|x\|\geq |f(x)|$ by definition of the norm. The desired inequality is then $\|f\|\|x_0\|\geq |f(x_0)|$ where $M=\|x_0\|$.
The direction $2 \implies 1$ is not that trivial however, any ideas to proceed?
We assume that 2 holds.
Consider the map $\psi:X\to\mathbb R^n$ given by $$ \psi(x)=(f_1(x),\ldots,f_n(x)). $$ Suppose that $(c_1,\ldots,c_n)$ is not in the image of $\psi$. As $\psi$ is linear, its image is a subspace of $\mathbb R^n$. As this subspace is proper, there exists a linear functional $\varphi$ on $\mathbb R^n$ such that $\varphi(c_1,\ldots,c_n)=1$ and $\varphi(f_1(x),\ldots,f_n(x))=0$ for all $x\in X$.
Linear functionals on $\mathbb R^n$ are of the form $(b_1,\ldots,b_n)\longmapsto \sum_ja_jb_j$. So there exist $a_1,\ldots,a_n\in\mathbb R$ such that $$\varphi(c_1,\ldots,c_n)=\sum_ja_jc_j=1$$ and $$\varphi(f_1(x),\ldots,f_n(x))=\sum_ja_jf_j(x)=0$$ for all $x\in X$. This contradicts 2, showing that $\psi$ must by surjective. In particular, there exists $x\in X$ with $(f_1(x),\ldots,f_n(x))=(c_1,\ldots,c_n)$.