Answer
see the proof below.
Work Step by Step
(a)
\begin{array}{l}{\quad\langle\mathbf{v}, \mathbf{v}\rangle= c_{1} v_{1} v_{1}+c_{2} v_{2} v_{2}+\cdots+c_{n} v_{n} v_{n}=c_{1} v_{1}^{2}+c_{2} v_{2}^{2}+\ldots+c_{n} v_{n}^{2} \geq 0} \\ {\text { Moreover, } c_{1} v_{1}^{2}+c_{2} v_{2}^{2}+\ldots+c_{n} v_{n}^{2} \text { is equal to zero only if } v_{1}=v_{2}=\cdots=} \\ {v_{n}=0, \text { or } \mathbf{v}=0}\end{array}
(b)
\begin{aligned}\langle\mathbf{u}, \mathbf{v}\rangle &= c_{1} u_{1} v_{1}+c_{2} u_{2} v_{2}+\cdots+c_{n} u_{n} v_{n} \\ &=c_{1} v_{1} u_{1}+c_{2} v_{2} u_{2}+\ldots+c_{n} v_{n} u_{n} \\ &=(\mathbf{v}, \mathbf{u}) \end{aligned}
(c)
\begin{aligned} k\langle\mathbf{u}, \mathbf{v}\rangle &= k\left(c_{1} u_{1} v_{1}+c_{2} u_{2} v_{2}+\cdots+c_{n} u_{n} v_{n}\right) \\ &=k c_{1} u_{1} v_{1}+k c_{2} u_{2} v_{2}+\cdots+k c_{n} u_{n} v_{n} \\ &=\langle c \mathbf{u}, \mathbf{v}\rangle \end{aligned}
(d)
\begin{aligned}\langle\mathbf{u}, \mathbf{v}+\mathbf{w}\rangle &= c_{1} u_{1}\left(v_{1}+w_{1}\right)+c_{2} u_{2}\left(v_{2}+w_{2}\right)+\cdots+c_{n} u_{n}\left(v_{n}+w_{n}\right) \\ &=c_{1} u_{1} v_{1}+c_{1} u_{1} w_{1}+c_{2} u_{2} v_{2}+c_{2} u_{2} w_{2}+\ldots+c_{n} u_{n} v_{n}+c_{n} u_{n} w_{n} \\ &=c_{1} u_{1} v_{1}+c_{2} u_{2} v_{2}+\cdots+c_{n} u_{n} v_{n}+c_{1} u_{1} w_{1}+c_{2} u_{2} w_{2}+\cdots+c_{n} u_{n} w_{n} \\ &=\langle\mathbf{u}, \mathbf{v}\rangle+\langle\mathbf{u}, \mathbf{w}\rangle \end{aligned}
