$V$ is a $n$-dimensional Euclidean space with an inner product denoted by $(\quad,\quad)$. $\{{\alpha}_{1},{\alpha }_{2},\cdots,{\alpha}_{n}\}$ is a basis of $V.$ $({\alpha}_{i},{\alpha }_{j})\leq 0,i\ne j.$
Show that
1.
For every nonzero vector $\alpha=\sum_{i=1}^{n}{a}_{i}{\alpha}_{i}$ ,if $({\alpha}_{i},\alpha)\geq 0$ for every $i\in \{1,2,\cdots,n\}$ ,then $a_{i}\geq 0$ for every $i\in \{1,2,\cdots,n\}.$
2.
Let two different nonzero vectors
$\alpha=\sum_{i=1}^{n}{a}_{i}{\alpha}_{i}，({\alpha}_{i},\alpha)\geq 0 $ for every $i\in \{1,2,\cdots,n\}$;
$\beta=\sum_{i=1}^{n}{b}_{i}{\alpha}_{i}$ , $({\alpha}_{i},\beta)\geq 0$ for every $i\in \{1,2,\cdots,n\}.$
If $\gamma= \sum_{i=1}^{n}{c}_{i}{\alpha}_{i} ,c_{i}=min\{a_{i},b_{i}\},$ then $({\alpha}_{i},\gamma)\geq 0$ for every $i\in \{1,2,\cdots,n\}.$
I want to prove 1. by induction on dimension $n$,but it seems complex to do so.
I have a proof here using induction. We first prove nothing catastrophic is happening (no induction for this step). Then use that to make a very clean induction using a well chosen subspace.
First we show that some coefficient must be positive.
Suppose all the coefficients were negative. Then we have $(\alpha_i,\alpha)\geq 0$ so $(a_i\alpha_i,\alpha)\leq 0$ for each $i$. Summing over $i$, we get $$ (\sum a_i\alpha_i,\alpha)=(\alpha,\alpha)\leq 0$$ But an inner product of a vector by itself must be non negative by definition of inner product. So $\alpha$ must be $0$, but this is a contradiction.
Now onto the induction.
0 (and 1) dimensional space work trivially.
Assume it works for $n-1$ dimensional space. Furthermore, assume that $0\leq(\alpha_i, \alpha)$ for all $1\leq i\leq n$. We want to show each $a_i\geq 0$.
We already know that for some $j$ between $1$ and $n$ we have $a_j\geq 0$. Now we note that because $B=\{\alpha_1,...\alpha_n\}$ forms a basis, the set $B\setminus \{\alpha_j\}$ forms a basis for an $n-1$ dimensional subspace. Furthermore, we know that $\beta=\alpha -a_j\alpha_j$ is in this subspace. We want to show that $\beta$ and $B\setminus \{\alpha_j\}$ satisfy the inductive hypothesis.
Let $i\neq j$ we have $$(\alpha_i,\alpha -a_j\alpha_j)= (\alpha_i,\alpha) -a_j(\alpha_i,\alpha_j) \geq 0$$
So we satisfy the inductive hypothesis, so we know that for all $i\neq j$ we have $a_i\geq 0$. And we assumed that $a_j$ is positive.
And we're done!