Prove a certain property of linear functionals, using the Hahn-Banach-Separation theorems I've been trying to solve this question, with no luck so far:
Let $X$ be a real linear space, and $\{\|\cdot \|_i\}_{i=1}^{n}$ family of norms on $X$. Let $f$ be a linear functional on $X$ such that for every $x\in X$: $f(x)\leq \max \|x\|_i$.
Prove that there are $t_1,...,t_n\geq0$ such that $\sum t_i=1$ and for every $x\in X$: $f(x)\leq\sum t_i \|x\|_i$.
I've been told that the best way to look at this questions is by separating $A=\mathop{\rm conv}\{(\|x\|_1-f(x),\ldots,\|x\|_n-f(x)):x\in X\}\subset \mathbb R^n$ with some other set.
It didn't do me much good to try this my own. My observations are that $A$ is obviously convex, and every point in $\{(\|x\|_1-f(x),\ldots,\|x|\|_n-f(x)):x\in X\}\$$ has at least one non-negative coordinate. $A$ isn't necessarily closed since $f$ isn't necessarily bounded with one specific norm (therefore not continuous), nor is opened. Don't know what separation theorem would do, and more than that - even if I could separate, all I'd get is $\varphi(A)\geq \gamma \geq \varphi(B)$ which would give me information about $\varphi(a)$ and not on $f$.
I thought it might be possible to prove that every point in a has at least one non-negative coordinate (is this even true?) and seperate from $\{(b_1,...,b_n):\forall i\:: b_i<0\}$
I'd love some tips/hints/guide/any thoughts actually.
Thanks a lot! 
 A: Feel free to read chunks of this answer until it gives you some new idea, then try to proceed on your own.
Notice that the thesis is the existence of $t_1,\dots,t_n\ge 0$ with $\sum t_i=1$ and such that $\sum t_i(\|x\|_i-f(x))\ge 0$. 
Your idea of choosing $B:=\{(b_1,\dots,b_n):\forall i\ b_i<0\}$ is good: now $A$ and $B$ are two convex sets and
if you can find a nontrivial functional $\phi$ which separates them (which means $\phi(a)\ge\phi(b)$ for any $a\in A$, $b\in B$) then you are done. 
In fact, any nontrivial functional $\phi:\mathbb{R}^n\to\mathbb{R}$ has the form $\phi(x)=\sum_{i=1}^n t_i x_i$ (where not all the $t_i$'s are $0$),
so in your case, since $0\in\overline{B}$, you get $\sum t_i(\|x\|_i-f(x))=\phi(a)\ge\phi(0)=0$ (where we have chosen $a:=(\|x\|_i-f(x))_i\in A$). 
Moreover, for any $i$, the vector $b:=(0,\dots,0,-1,0,\dots,0)$ having $-1$ as the $i$-th coordinate lies in $\overline{B}$, while $0\in A$ (just choose $x=0$!); thus
$-t_i=\phi(b)\le\phi(0)=0$ and you obtain $t_i\ge 0$. Finally by rescaling your functional by a positive constant you may suppose $\sum t_i=1$ and
the property $\forall x\ \sum t_i(\|x\|_i-f(x))\ge 0$ is obviously still true (it is invariant under rescaling), so you are finished! 
We are left to check the existence of a nontrivial functional $\phi$: here Hahn-Banach is your friend. You can verify the so-called first geometric form
(separation of an open convex set from a convex set): $B$ is open and convex, while $A$ is convex. If we show that $A\cap B=\emptyset$, then all the hypotheses of this form of Hahn-Banach are satisfied and we are done. 
Notice that we have not yet used the fact that $\|\cdot\|_i$ are norms. Suppose $y\in A\cap B$: as $y\in A$ there exist $\lambda_1,\dots,\lambda_m\ge 0$
such that $\sum_{j=1}^m \lambda_j=1$ and some $x_1,\dots,x_m\in X$ such that $y=\sum_{j=1}^m\lambda_j(\|x_j\|_i-f(x_j))_{i=1}^n$. 
(By $(\|x_j\|_i-f(x_j))_{i=1}^n$ I mean the vector $(\|x_j\|_1-f(x_j),\dots,\|x_j\|_n-f(x_j))$ ). 
So componentwise we have
$$\forall i\ \sum_{j=1}^m\lambda_j(\|x_j\|_i-f(x_j))=y_i<0$$ (since $y\in B$). So
$$\forall i\ f(\sum \lambda_j x_j)=\sum \lambda_j f(x_j)>\sum\lambda_j\|x_j\|_i\ge\|\sum \lambda_j x_j\|_i$$
(the last inequality comes from the properties of a norm) and if we put $x:=\sum \lambda_j x_j$ this chain of inequalities gives
$\forall i\ f(x)>\|x\|_i$, which contradicts the hypothesis $f(x)\le\max\|x\|_i$.
This contradiction proves that $A\cap B=\emptyset$.
