$t\in (0,1)$ and $f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$. Show that if strict inequality holds for even one $t$, then it holds for all $t$. This is a part of a solution to a problem in showing that if $f$ is continuous and satisfies the condition $f([x+y]/2)\lt [f(x)+f(y)]/2$, then $f$ is convex.
Let $t\in (0,1)$. 
We have the weak inequality $f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$. And from a previous part of the problem, we know that the strict inequality holds for all $t$ of the form $m/2^n$.
Now the solution claims that if strict inequality holds for even one $t$, then it holds for all $t$ (by applying the weak inequality to $x$ and $tx+(1-t)y$ or to $tx+(1-t)y$ and $y$. Hence, we must have strict inequality for all $t$. 
However, I've been stuck on proving the above claim. I've tried to show it but I don't see how to apply the weak inequality to obtain the desired result. How can I show this? I would greatly appreciate any help.
 A: Fix $x$ and $y$, wlog $x<y$ (*). Suppose you have strict inequality for some $\sigma \in (0,1)$:
$$f(\sigma x + (1-\sigma)y) < \sigma f(x) + (1-\sigma )f(y)$$
Now assume $t\in (0,1)$, wlog $0< t <\sigma $, the other case ($\sigma < t < 1$) is, of course, similar. 
Denote $z:=\sigma x + (1-\sigma)y$. Note that then $z < tx + (1-t)y =: w < y$ (just look at the case $t =0 $ to verify this). So there is $\alpha \in (0,1)$ such that $w = \alpha z + (1-\alpha)y$, and therefore
$$f(w) \le \alpha f(z) + (1-\alpha) f(y) < \alpha \left\{\sigma f(x) + (1-\sigma)f(y)\right\} + (1-\alpha)f(y) \\ = \alpha\sigma f(x) + \left\{\alpha(1-\sigma) + (1-\alpha)\right\}f(y)  = \alpha\sigma f(x) + (1-\alpha \sigma)f(y)$$
Note the strict inequality. 
Now also note that by definition of $z$
$$w=  \alpha z + (1-\alpha)y = \alpha(\sigma x + (1-\sigma)y)+ (1-\alpha)y = \alpha\sigma x + \left(1-\alpha\sigma \right)y$$
so, by strict monotonicity of $t\mapsto tx + (1-t)y$ and definition of $w$ we must have $\alpha \sigma = t$. So the last inequality is exactly what was claimed (just plug in $w$ in $f(w)$ on the lhs and $t=\alpha \sigma $ on the rhs). 
(*) you don't need $x$ and $y$ to be real numbers, only for convenience the direction from $x$ to $y$ is distinguished, so later on we can say that $w$ lies between $z$ and $y$.
