Let $x,y\in\mathbb{R}^n$ be such that $||x+ty||\geq||x||$ $\forall \ t \in \mathbb{R}$. Show that $\vec{x} \cdot \vec{y} = 0$ Let x,y$\in\mathbb{R}^n$ be such that
     $||x+ty||\geq||x||$
  $\forall $  $t  \in \mathbb{R}$
Show that x.y=0
Approach.
I expanded the above inequality then I get
$-2t$ x.y $\leq$ $t^2||y||^2$
$\forall $  $t  \in \mathbb{R}$
Now If v=0 then x.y=0
AND 
if v$\neq0$ then let $t=\frac{x.y}{||v||^2}$
Putting into the simplified inequality we get $-2$x.y  $\leq$ x.y
Hence
$3$x.y$\leq0$
Which implies  x.y$=0$
Is it correct?
Correct me if not.
Someone told me that it also holds for general vector space too.How do I prove it.I am unable to do anuthing except expanding the inequality.
Thanks
 A: Let
$$
f(t)=\|x+ty\|^2-\|x\|^2=t^2\|y\|^2+2tx\cdot y
$$
Since $f(0)=0$, if $f(t)\ge0$ for all $t$ in a neighborhood of $0$, we must have $f'(0)=0$.
This means $2x\cdot y=0$.
A: We want to prove that 
$$
\|x\| \leq \|x + ty\| \quad \forall t \in \mathbb{R} \implies \langle x, y \rangle = 0.
$$
We prove the contrapositive, namely
$$
\langle x, y \rangle \neq 0 \implies \text{there exists } t \in \mathbb{R} \text{ such that } \|x\| \leq \|x + ty\| \text{ is false}.
$$
As you did, rewrite the inequality as 
$$
0 \leq 2t \langle x, y \rangle + t^2 \|y\|^2.
$$
Since we're assuming $\langle x, y \rangle \neq 0$ for the contrapositive, and we only need to find one value of $t \in \mathbb{R}$ which violates the given inequality, we may assume $t \neq 0$ and divide by $t \langle x, y \rangle \neq 0$ to get
$$
0 \leq 2 + t \frac{\|y\|^2}{\langle x, y \rangle}.
$$
Now, if $\langle x, y \rangle > 0$, the inequality is clearly false for $t \ll 0$. Similarly, if $\langle x, y \rangle < 0$, the inequality is clearly false for $t \gg 0$. 
A: For $||y||= 0$ there is nothing to show. So, let $||y|| \neq 0$.
According to our assumption we have $\forall t \in \mathbb{R}$:
$$||x + ty||^2 = ||x||^2+ t^2||y||^2 + 2t (x\cdot y) \geq ||x||^2 $$
Rearranging and square completion gives:
$$(t||y||- \frac{x\cdot y}{||y||} )^2 \geq \frac{(x\cdot y)^2}{||y||^2} \stackrel{t=\frac{x\cdot y}{||y||^2}}{\longrightarrow}x \cdot y = 0$$
