Answer
see the proof below.
Work Step by Step
For any $(u_1,u_2), (v_1, v_2 ), (w_1, w_2) \in {R}^2 , k \in {R}$
(1) $\langle (u_1,u_2), (u_1, u_2 )\rangle=\frac{1}{2}u_1^2+\frac{1}{4}u_2^2>0$ and $\langle (u_1,u_2), (u_1, u_2 )\rangle=\frac{1}{2}u_1^2+\frac{1}{4}u_2^2=0$ if and only if $u_1=0$, $u_2=0$.
(2) \begin{align*}
\langle (u_1,u_2), (v_1, v_2 )\rangle&=\frac{1}{2}u_1v_1+\frac{1}{4}u_2v_2\\
&=\frac{1}{2}v_1u_1+\frac{1}{4}v_2u_2\\
&=\langle (v_1, v_2 ),(u_1,u_2)\rangle.
\end{align*}
(3) \begin{align*}
\langle (ku_1,ku_2), (v_1, v_2 )\rangle&=\frac{1}{2}ku_1v_1+\frac{1}{4}ku_2v_2\\
&=k(\frac{1}{2}u_1v_1+\frac{1}{4}u_2v_2)\\
&=k\langle (u_1,u_2), (v_1, v_2 )\rangle.
\end{align*}
(4) \begin{align*}
\langle (u_1,u_2)+(v_1, v_2 ), (w_1, w_2)\rangle&=\langle (u_1+v_1, u_2+ v_2 ),(w_1, w_2)\rangle\\
&=\frac{1}{2}(u_1+v_1)w_1+\frac{1}{4}(u_2+ v_2)w_2\\
&=\frac{1}{2}u_1w_1+\frac{1}{2}v_1w_1+\frac{1}{4}u_2w_2+\frac{1}{4} v_2w_2\\
&=(\frac{1}{2}u_1w_1+\frac{1}{4}u_2w_2)+(\frac{1}{2}v_1w_1+ \frac{1}{4}v_2w_2)\\
&=\langle (u_1, u_2 ),(w_1,w_2)\rangle+\langle (v_1, v_2 ),(w_1,w_2)\rangle.
\end{align*}
