Strict inequality in a Hilbert Space 
Let $H$ be a Hilbert Space and let $x$ and $y$ be two distinct points in $H$ such that $||x|| = ||y|| = 1$. Then show that $||tx+(1-t)y||<1$, where $0<t<1$.

It is pretty clear that any point lying on the line joining two distinct points on a sphere which is not an end-point itself will lie inside the sphere. But I have problem deducing the strict inequality here. We see that:
$$\begin{align} ||tx+(1-t)y||^2 = \langle tx+(1-t)y, tx+(1-t)y\rangle =  t^2||x||^2+2t(1-t)Re\langle x,y\rangle +(1-t)^2||y||^2  \leq (t+(1-t))^2 = 1 \end{align}$$.
So, my question is how to get rid of the equality. I can not argue that $x$ and $y$ are linearly independent as these may be diametrically opposite points (on the sphere) and hence linearly dependent.
 A: This is actually a problem about inner products.
The last inequality is actually Cauchy-Schwarz inequality.

$$|\langle x,y\rangle| \le ||x|| \, ||y||$$
  Equality holds iff $x$ and $y$ are dependent.

However, the given conditions that $x \ne y$ and $||x||=||y||=1$ leave us two possibilities:


*

*$x$ and $y$ are linearly independent, so
$$\Re\langle x,y\rangle \le |\langle x,y\rangle| < ||x|| \, ||y|| = 1.$$
Put this strict inequality into to your step, and you get your desired conclusion.

*$y = kx$:  $||y|| = 1 \implies |k| = 1$


*

*in $\Bbb{R}$, that's $x$ and $y$ are on the opposite poles of a unit sphere.  Use the given condition $t \in (0,1)$ to finish this proof.

*in $\Bbb{C}$, $\Re\langle x,y\rangle=\Re\left(\bar{k}\right) = \Re(k) \le 1$.  Use $x \ne y$ to make the previous inequality strict:
$$||x-y|| = ||(k-1)x|| = |k-1| \ne 0 \implies k \ne 1$$
Put $\Re\langle x,y\rangle = \Re(k) < 1$ back into your proof, and we're done.



Remarks: When you are stuck, think of which condition(s) is(are) unused.
