Lindenstrauss defines in his paper that a Banach space $X$ is strictly convex if every boundary point of unit ball in $X$ is an exposed point. (A point of $B_X$(unit ball) is called an exposed point if there is $f\in X^*$ such that $f(x)>f(y) \forall y\neq x$ in $B_X$)
I already know that X is strictly convex if $\Vert x\Vert=\Vert y\Vert=1$ implies $\Vert x+y\Vert<2$, so they should be equivalent.
I can show the necessity : set $f(x)=\Vert x\Vert$ for $\Vert f\Vert=1$ and if $f(x)=f(y)$ that it violates the condition $\Vert x+y\Vert<2$.
But how to show sufficiency?
Both conditions are equivalent to: the unit sphere does not contain a nontrivial line segment.
Indeed, the existence of $x\ne y$ with $\|x\|=\|y\|=\|(x+y)/2\|=1$ means precisely that the line segment $[x,y]$ lies on the unit sphere.
If $[x,y]$ lies on the unit sphere, then $z=(x+y)/2$ is not exposed since $$f(z)=(f(x)+f(y))/2 \le \max(f(x), f(y))$$ Conversely, if $z$ is not exposed, then let $f$ be a norming functional for $z$ and observe that the set $\{x: \|x\|\le 1, f(x)=1 \}$ is convex, has more than one point, and lies on the unit sphere.