Let $H$ be an inner product space and $u,u_{1},u_{2}\in H$, show that $$\|u-u_{1}\|^{2}+\|u-u_{2}\|^{2}$$ obtains minimum when $u=\dfrac{u_{1}+u_{2}}{2}$.
I begin my solution by using the reversed triangle inequality, i.e. $$\|u-u_{1}\|^{2}+\|u-u_{2}\|^{2}\geq2\|u\|^{2}+\|u_{1}\|^{2}+\|u_{2}\|^{2}-2\|u\|(\|u_{1}\|+\|u_{1}\|).$$
I have tried using other inequalities but they don't seem to make progress. How do I proceed from here?
On
Maybe try it the other way around: If you insert $u_{min}=\frac{u_1+u_2}{2}$, you obtain for your expression $f(u) = \|u_1-u\|^2 + \|u_2-u\|^2=2\|\frac{u_1}{2}-\frac{u_2}{2}\|^2 = \frac{1}{2} \|u_1-u_2\|^2$.
Now to show that this is a minimum, you can see that $ \frac{1}{2} \|u_1-u_2\|^2 \leq f(u)$ for all $u$ by the triangle inequality. That goes like this:
$$ \frac{1}{2} \|u_1-u_2\|^2 \leq \frac{1}{2} \left( \|u_1-u\| + \|u_2-u\| \right)^2 \leq \|u_1-u\|^2 + \|u_2-u\|^2 = f(u)$$ The last step is by the general inequality $2ab \leq a^2 + b^2$ valid for all real numbers.
If you want to show that this is a global minimum, some additional work will have to be done, but maybe this here already helps to look in the right direction.
$$\Vert u-u_1\Vert^2+\Vert u-u_2\Vert^2=\Vert u-u_1\Vert^2+\Vert u_2-u\Vert^2\geq\frac{1}{2}\Vert u-u_1+u_2-u\Vert^2=\frac{1}{2}\Vert u_2-u_1\Vert^2.$$ The equality occurs for $$u-u_1=u_2-u.$$