An inner product on a vector space $V$ assigns to vectors $\mathbf{u}, \mathbf{v}$ a real number $(\mathbf{u}, \mathbf{v})$, such that
(1) $(\mathbf{u}, \mathbf{u}) \geq 0$ for all $\mathbf{u}$, and $(\mathbf{u}, \mathbf{u})=0$ if and only if $\mathbf{u}=\mathbf{0}$
(2) $(\mathbf{u}, \mathbf{v})=(\mathbf{v}, \mathbf{u})$ for all $\mathbf{u}, \mathbf{v} ;$
(3) $(\mathbf{u}, a \mathbf{v}+b \mathbf{w})=a(\mathbf{u}, \mathbf{v})+b(\mathbf{u}, \mathbf{w})$ for all $\mathbf{u},\mathbf{v}, \mathbf{w}$
(4)Norm of a vector $\mathbf u$ is $(\mathbf{u}, \mathbf{u})$ ,
Example : Let $V=\mathbb{R}^{2}, \mathbf{u}=\left[\begin{array}{l}u_{1} \\ u_{2}\end{array}\right], \mathbf{v}=\left[\begin{array}{l}v_{1} \\ v_{2}\end{array}\right] .$ Define $(\mathbf{u}, \mathbf{v})=2 u_{1} v_{1}-u_{1} v_{2}-v_{1} u_{2}+u_{2} v_{2}$
This is an inner product with strange norm for its vectors.
Is it always possible to draw a triangle with its lengths equal to the norms of $\mathbf u,\mathbf v,\mathbf {v-u}$ no matter how contrived the inner product and hence the norm is?