Unit vectors in $\Bbb R^n$ Suppose that $x$ and $y$ are unit vectors in $\Bbb R^n$. Show that if $\left\Vert{{x+y}\over2}\right\Vert=1$ , then $x =y$.
Please enlighten me for this problem!
Thanks in advance!
 A: $\|x\|=\|y\|=1$ and $\|x+y\|=2$. Therefore,
$$
\begin{align}
4
&=\|x+y\|^2\\
&=\|x\|^2+2x\cdot y+\|y\|^2\\
&=2+2x\cdot y
\end{align}
$$
so that $x\cdot y=1$. Thus,
$$
\begin{align}
\|x-y\|^2
&=\|x\|^2-2x\cdot y+\|y\|^2\\
&=0
\end{align}
$$
Therefore, $x=y$.
A: Do you know the triangle inequality? Norms satisfy the triangle inequality $\|u+v\| \le \|u\|+\|v\|$.
To prove what you have written, just note that $$\|x\|=\|x+(y-x)\| \le \|x\|+\|y-x\|=\|x\|+\|x-y\|.$$
This holds for any vectors in $\Bbb R^n$, not just unit vectors.
A: The condition gives $\|x+y\|=2=\|x\|+\|y\|$.    A part of the triangle inequality in Euclidean space is the fact that equality holds precisely when one of the two vectors is a multiple of the other.  Hence $x=ky$; but since each are unit vectors we must have $|k|=1$.  $k=-1$ doesn't work since then $\|x+y\|=0$, so $k=1$.
A: $$1=1^2=||\frac{x+y}{2}||^2=\frac{1}{4}<(x+y),(x+y)>$$
$$4=<x,x>+2<x,y>+<y,y>=2+2<x,y>$$
$$<x,y>=1$$ So x and y are the same unit vector. Otherwise $|<x,y>|<1$ or $<x,y>=-1$ if $y=-x$.
