find maximum length of sum of two vector let us consider following problem

so  let us introduce  vector with   $2$ coordinates ,namely
$v=(v_1,v_2)$ with length  $12$,which means that 
$v_1^2+v_2^2=144$
and vector $s=(s_1,s_2)$ with length   $10$,so it   means
$s_1^2+s_2^2=100$
now let us consider  sum of two vector
$v+s=(v_1+s_1,v_2+s_2)$
if we take  square  of each member ,we get
$v_1^2+2*v_1*s_1+s_1^2+v_2^2+2*v_2*s_2+s_2^2$
so  it means that
$244+2*v_1*s_1+2*v_2*s_2$
now   for vector  $v$, possible solutions is $(8,6)$ or $(6,8)$ ,for vector $s$  we may write $144$ what we could do?please help me
 A: You are making it too hard.
If the two vectors are $\mathbf{u}$ and $\mathbf{v}$, what does the triangle inequality say about the maximum length of $\mathbf{u}+\mathbf{v}$?
Once you have an upper bound, you need to show that it is tight. It won't be hard.
A: If you look at it from a geometrical point of view, it becomes clear that you can get the maximum length if both vectors point in the same direction. That way, you can simply add the length of the two vectors (the triangle formed by the two vectors is then completely stretched out).
A: I think we can consider the term $v_1s_1+v_2s_2$. In fact it is the product of two vectors:
$$v*s=v_1s_1+v_2s_2,$$ we have also$$v*s=||v||||s||cos(\theta)$$ where $\theta$ is the angle between the two vectors. So we have:$$v_1s_1+v_2s_2=v*s=||v||||s||cos(\theta)\leq||v||||s||=12*10=120$$
We have the maximum when $cos(\theta)=1$, so $244+2∗v_1∗s_1+2∗v_2∗s_2\leq244+2*120=484$, so the maximum of length is $\sqrt{484}=22$. In fact the sum of two vector is maximum when the two vectors have a same direction, that means that the angle between them is $0$.
