Answer
see the proof below.
Work Step by Step
For any $(u_1,u_2,u_3), (v_1, v_2,v_3 ), (w_1, w_2,w_3) \in {R}^3 , k \in {R}$
(1) $\langle (u_1,u_2,u_3), (u_1, u_2 ,u_3)\rangle= \frac{1}{2}u_1^2+ \frac{1}{4}u_2^2+\frac{1}{2}u_3^2>0$ and $\langle(u_1,u_2,u_3), (u_1, u_2 ,u_3)\rangle= \frac{1}{2}u_1^2+\frac{1}{4}u_2^2+\frac{1}{2}u_3^2=0$ if and only if $u_1=0$, $u_2=0$, $u_3=0$.
(2) \begin{align*}
\langle (u_1,u_2,u_3), (v_1, v_2,v_3 )\rangle&= \frac{1}{2}u_1v_1+\frac{1}{4}u_2v_2+\frac{1}{2}u_3v_3\\
&=\frac{1}{2} v_1u_1+\frac{1}{4}v_2u_2+\frac{1}{2}v_3u_3\\
&=\langle (v_1, v_2,v_3 ),(u_1,u_2,u_3)\rangle.
\end{align*}
(3) \begin{align*}
\langle (ku_1,ku_2,ku_3), (v_1, v_2 ,v_3)\rangle &=\frac{1}{2}ku_1v_1+\frac{1}{4}ku_2v_2+\frac{1}{2}ku_3v_3\\
&=k(\frac{1}{2}u_1v_1+\frac{1}{4}u_2v_2+\frac{1}{2}u_3v_3)\\
&=k\langle(u_1,u_2,u_3), (v_1, v_2,v_3 )\rangle.
\end{align*}
(4)
\begin{aligned}
&\left\langle\left(u_{1}, u_{2}, u_{3}\right)+\left(v_{1}, v_{2}, v_{3}\right),\left(w_{1}, w_{2}, w_{3}\right)\right\rangle =\left\langle\left(u_{1}+v_{1}, u_{2}+v_{2}, u_{3}+v_{3}\right),\left(w_{1}, w_{2}, w_{3}\right)\right\rangle \\ &=\frac{1}{2}\left(u_{1}+v_{1}\right) w_{1}+\frac{1}{4}\left(u_{2}+v_{2}\right) w_{2}+\frac{1}{2}\left(u_{3}+v_{3}\right) w_{3} \\ &=\left(\frac{1}{2} u_{1} w_{1}+\frac{1}{4} u_{2} w_{2}+\frac{1}{2} u_{3} w_{3}\right)+\left(\frac{1}{2} v_{1} w_{1}+\frac{1}{4} v_{2} w_{2}+\frac{1}{2} v_{3} w_{3}\right) \\ &=\left\langle\left(u_{1}, u_{2}, u_{3}\right),\left(w_{1}, w_{2}, w_{3}\right)\right\rangle+\left\langle\left(v_{1}, v_{2}, v_{3}\right),\left(w_{1}, w_{2}, w_{3}\right)\right\rangle \end{aligned}
