Proving the inequality of Cauchy-Schwarz in an Euclidean space. It says let (G, <.,.>) be an euclidean space. Show that for all x, y belonging to G: 
modulus<x,y> <= sqrt<x,x> * sqrt<y,y>

and in the mark scheme they put:
for all x and  y and a real C:
0 <=  =  -2C + C^2. 
if y = 0, there is nothing to prove (the two members of the inequality are 0). if y =/= 0, then  > 0 (this is the only part i get lol). the minimum of this polynomial of degree 2 in C is found by C = /. We get, 
0 <=  - (^2)/ 
apologies for the terrible writing i literally translated it as it was in french. i dont understand how we're supposed to know that we are supposed to use a constant C to prove it. I am really bad with proofs, can someone explain why they did what they did in the mark scheme?
Thanks!!!!
 A: The idea of the proof is the following: we want an inequality, so let's take the simplest one, that the inner product of something with itself is always positive. But, that won't be enough, so we do a little trick. See...Consider $x, y \in G$, and a constant $C$. Then, we have: $$\langle x + Cy, x + Cy \rangle \geq 0 \\ \langle x,x \rangle + 2C \langle x, y \rangle + C^2 \langle y, y \rangle \geq 0$$
This last expression is a quadratic polynomial in $C$, and since it's always positive it can have at most one zero. Therefore we know it has $\Delta \le 0$ (because $\Delta \gt 0$ would give two zeros), that is: $$\left(2 \langle x, y \rangle \right)^2 - 4 \langle y, y \rangle \langle x, x \rangle \leq 0 \\ 4 \langle x, y \rangle^2 \leq 4\langle y, y \rangle \langle x, x \rangle\\\vert \langle x, y\rangle \vert \leq \vert \vert x \vert \vert \cdot \vert \vert y \vert \vert$$
Since $\langle x, x \rangle = \vert \vert x \vert \vert^2$, and we take the root on both sides of the expression. I hope this helps!
