Proof of an equivalent definition of strictly convex? $X$ 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?  

$$\|(x+y)/2\|< 1 \implies \|x+y\|<2\tag1\label1$$
$$ax+(1-a)y = a(x+y)+(1-2a)y\tag2\label2$$
$$ax+(1-a)y = (1-a)(x+y)+(2a-1)x\tag3\label3$$
Break it into cases:
Case $1$: $a\leq1/2$
Use the triangle inequality on $\eqref2$ and substitute $\eqref1$.
Case $2$: $a\geq1/2$
Use triangle inequality on $\eqref3$ and substitute $\eqref1$.
Note: $\|cx\| = c\|x\| \iff c\geq0$.
