Innerproduct of <a+b, c> = <c, c> where c != 0

Question

Say we have 3 non zero vectors a,b,c. If (a + b, c) = (c, c), can we conclude that a + b = c?

Here is my attempt to prove the claim: a + b != 0, since (a + b, c) is non zero.

I tried subtracting (c , c) from each side:

(a + b, c) - (c ,c) = (c , c) - (c , c)

(a + b - c,c) = 0 -> a + b-c = 0 -> a + b = c.

Is the proof correct? edit: The proof is not correct since (a+b-c,c) = 0 does not imply that a+b-c = 0.

Answer 1

Equality of two inner products does not imply equality of the vectors. Take $c=(1,1,-1)^T$. Then $\langle c,c\rangle =3$. Let $a+b=(0,0,-3)^T$. Then again $\langle a+b,c\rangle=3$, but $a+b\neq c$.