Suppose $f:V\times V \rightarrow \mathbb R$ is a symmetric indefinite bilinearform. Show $\exists v\neq0:f(v,v)=0$ Suppose $f:V\times V \rightarrow \mathbb R$ is a symmetric indefinite bilinearform.
Show $\exists v\neq0:f(v,v)=0$
Proof: Since $f$ indefinite $\exists x,y \in V:f(x,x)>0 \& f(y,y)<0$.
Wlog $f(x,x)=1$ and $f(y,y)=-1$
Now we gonna apply Gram Schmidt process on both of them.
$\bar y:=y-\frac{f(y,x)}{f(x,x)}x=y-f(x,y)x$
$f(\bar y,\bar y)=f(y,y)+f(x,y)^2-2f(x,y)f(y,x)=-1-f(x,y)^2<0$
$y':=\frac{\bar y}{\sqrt{|f(\bar y,\bar y)|}}$   then $f(y',y')=-1$ and $x\perp y'$.
$f(x+y',x+y')=f(x,x)+f(y',y')+2f(x,y')=1+-1+2\cdot 0=0$
qed
Is the proof correct?
I would like to see your creative approaches.
 A: In case you only care about the existence, you may also do it slightly differently. Define
$$ g(t)= f(tx + (1-t)y, tx + (1-t)y)= t^2 f(x,x) + 2 t(1-t) f(x,y) + (1-t)^2 f(y,y), $$
where $x,y$ are as in your text.
This function is continuous and $g(1)=f(x,x)>0$ and $g(0)=f(y,y)<0$ and by the intermediate value theorem you are done. In fact we also need to check that the resulting vector is not the zero vector, however, this would only be possible if $x=\alpha y$ with $\alpha <0$, but then $f(x,x)$ and $f(y,y)$ had the same sign. 
Note that this is essentially the same thing you did, but just in a lazy way. Furthermore, this argument does not use that we are in a linear algebra setting, i.e. we are not using that $V$ is finite-dimensional (in the finite-dimensional setting you would not even need to write the second equality in the definition of $g$ to see that it is continuous, as bilinear maps on finite-dimensional normed spaces are continuous).
A: Let $x,y$ be as in your solution.
For $r \in \mathbb{R}$ we have
$$f(rx+y, rx+y) = f(x,x)r^2 + 2f(x,y)r + f(y,y)$$
This quadratic in $r$ has a real root $r_0$ because the discriminant is
$$4\underbrace{f(x,y)^2}_{\ge 0} - 4\underbrace{f(x,x)}_{>0}\underbrace{f(y,y)}_{<0} > 0$$
Hence $f(r_0x+y, r_0x+y) = 0$.
A: Here is another proof came to my mind.       
Since $f$ symmetric and $2$ invertible, it exists an orthogonal basis $(b_1,...,b_n)$.
Inertia law assures $\exists i\neq j :\  f(b_i,b_i)=c_i>0;  f(b_j,b_j)=c_j<0$,since $f$ is indefinite.
Again WLOG we asume $c_i=1, c_j=-1$
and get $f(b_i+b_j,b_i+b_j)=0$.
