Relation of angle between $e_1,e_2$ with angle between $e_1^*, e_2^*$ in Euclidenan space $\mathbb{R}^2$ Consider Euclidean space $\mathbb{R}^2$. 

Let $\{e_1,e_2\}$ be an ordered basis with angle between them acute (i.e. $(e_1,e_2)\leq 0$. Let $\{f_1,f_2\}$ be dual of above basis w.r.t. inner product $(\cdot,\cdot)$. I want to show that $(f_1,f_1)\geq 0$. 

Let $f_1=ae_1+be_2$ and $f_2=ce_1+de_2$. Now $(f_i,e_j)=\delta_{ij}$ so 
$$a=(f_1,f_1), \hskip5mm b=(f_1,f_2), \hskip5mm c=(f_2,f_1),\hskip5mm d=(f_2,f_2).$$
Then $(f_1,f_2) = (ae_1+be_2, ce_1+de_2)= ac|e_1|^2 + bd |e_2|^2+(ad+bc)(e_1,e_2).$
For simplicity we can assume that $e_1,e_2$ are unit vectors. So 
$$(f_1,f_2) =|f_1|^2.(f_2,f_1) +(f_1,f_2).|f_2|^2+\{ |f_1|^2.|f_2|^2+(f_1,f_2)^2\} (e_1,e_2)$$
i.e. 
$$\{1-|f_1|^2-|f_2|^2\} (f_1,f_2) =\{ |f_1|^2.|f_2|^2+(f_1,f_2)^2\} (e_1,e_2).$$
After this, I couldn't proceed, and I don't know if this is the way to proceed for proof. How can we prove the assertion?
 A: (1) Let $A=[a_{ij}]=[(e_i,e_j)]$ be the Gram matrix of basis $(e_i)_i$, and $B=[b_{ij}]=[(f_i,f_j)]$ the Gram matrix of dual basis $(f_i)_i$ of $(e_i)_i$ w.r.t. $(\cdot,\cdot)$.
(2) Then show that $[b_{ij}]=[a_{ij}]^{-1}$.
(3) Consider the above case for basis $e_1,e_2$. Show that (1,2)-th entry of $B$ and (1,2)-th entry of $A$ are related by 
$$(f_1,f_2) = \frac{-1}{(e_1,e_1).(e_2,e_2)-(e_1,e_2)^2}(e_1,e_2).$$
(4) By Cauchy-Schwartz, $(v_1,v_2)^2\leq |v_1|^2.|v_2|^2$ with equality only when $v_1,v_2$ are dependent.
(5) Since $e_1,e_2$ are independent, the denominator of above equation is positive.
(6) q.e.d.
A: I think we can use the natural geometric proof. Let $\alpha$ be the angle between $e_1$ and $e_2$ and $\beta$ be the angle between $f_1$ and $f_2$. It's enough to show that $\alpha+\beta=\pi$ i.e. if $\alpha$ is obtuse then $\beta$ is acute. We can see that $f_1$ is orthogonal to $e_2$ and $f_2$ is orthogonal to $e_1$. Also, $f_i$ and $e_i$ should have an acute angle since $$(f_i,e_i)=|f_i||e_i|\cos(\theta_i)=1>0$$ i.e. $\cos(\theta_i)>0$. Then $\alpha+\beta=\pi$ follows from this picture.
