Proof of an equivalent definition of strictly convex?

Question

$X$ is a normed space. If for all $x,y\in X$ such that $\|x\|=\|y\|=1, x\neq y$, we have that $\|\frac{x+y}{2}\|<1$, then we know that $X$ is strictly convex. How can I show that for all $\lambda \in (0,1)$, $\|\lambda x + (1-\lambda) y\|<1$ always holds?

Answer 1

The point $\lambda x+ (1-\lambda) y$ is a convex combination of $\frac12(x+y)$ and of $x$ or $y$. Then the norm of $\lambda x+ (1-\lambda) y$ is, by convexity of the norm, $\le$ a convex combination of $\|\frac12(x+y)\|$ and $1$, which is strictly less that one.