Inequality for a differentiable concave function on $[0,1]$ Note: Concave refers to this definition.
Let $f: [0,1] \rightarrow \mathbb{R}$ be differentiable, concave and not identically $0$. Assume $f(0) = f(1) = 0.$ Let $a \in (0,\frac{1}{2})$. Show that 
$$f(y) > af(x), \forall x \in [0,1] \text{ and } \forall y\in [a,1-a].$$
My attempt:
By Rolle's theorem we know that there is a $c \in [0,1]$ such that $f'(c) = 0$. Then $c$ is a local extrema, and since $f$ is concave, it is the only extrema of the function, and it is a maximal point. Let $m = f(c) > 0$. 
Then it is enough to prove that $f(y) > am \hspace{0.2cm}\forall y \in [a, 1-a]$. Since $f$ is concave, $f$ achieves it's minimum on $[a,1-a]$ in $a$ or $1-a$, so it's enough to prove that 
$$\min\{f(a),f(1-a)\} > am \hspace{0.2cm} \forall a \in (0,\frac{1}{2})$$
This amounts to considering $g(x) = f(x) - xm$ and $h(x) = f(1 - x) - xm$ and showing that they are $> 0$ on $(0,\frac{1}{2})$.
For $g(x)$, for example, i tried differentiating it and looking for extrema: $g'(x) = f'(x) - m$. However I do not know what to say about the equation $f'(x) = m$, or about $f'(x)$ monotonicity (i guess that it is decreasing since $f$ is concave, but that does not seem to make sense when looking at a graph for instance). Any ideas about how to continue this? 
 A: Since $f$ is concave, $f$ satisfies 
$$
f( (1-\alpha)x_1 + \alpha x_2) \geq (1-\alpha)f(x_1)+\alpha f(x_2) \tag{1}
$$
for all $x_1, x_2 \in [0,1]$ and all $\alpha \in (0,1)$. 
To prove 
$$
f(y)> a f(x) \tag{2}
$$
for all $x \in [0,1]$ and $y \in [a, 1-a]$, we will need to use the fact that 
$f(t) > 0$ for all $t \in (0,1)$. The inequality $f(t) \geq 0$ is immediate from (1) by substituting $0$ for $x_1$ and $1$ for $x_2$; this gives
$$
f(t) = f( (1-t)(0) + t(1)) \geq (1-t)f(0) + t f(1) = 0.
$$ 
To upgrade this to strict inequality, suppose for contradiction that there were some $t_1 \in (0,1)$ with $f(t_1) = 0$.  Since $f \geq 0$ on $[0,1]$, it would follow that $t_1$ was a local minimum for $f$ and therefore that $f’(t_1) = 0$.
Assuming that $f$ is not identically $0$, there would be a $t_2$ in $(0,t_1) \cup (t_1, 1)$ with $f(t_2) > 0$.  Supposing WLOG that $t_2 \in (0, t_1)$, it would follow by the Mean Value Theorem that there was a $t_3 \in (t_2, t_1)$ so that $f’(t_3) = (t_1-t_2)(f(t_1) - f(t_2)) < 0 = f’(t_1)$, contradicting the fact that a differentiable concave function must have a derivative that is non-increasing.
Having proved that $f(t) > 0$ for all $t \in (0,1)$, we can prove (2) from (1) as follows:
writing
$$
y = ax + b = ax+(1-a)z,
$$
we will show that if $x \in (0,1)$, then $z$ is in $(0,1)$ so that we can substitute $x$ for $x_2$ and $z$ for $x_1$ in (1).
To give a lower bound for $z$, we use the inequality $y \leq 1-a$.  Assuming that $x > 0$, we have 
$$
z = \frac{y-ax}{1-a} \leq \frac{(1-a) - ax}{1-a} = 1 - \frac{ax}{1-a} < 1.
$$
To give a lower bound, use the inequality $y \geq ax$ along with the assumption $x < 1$ to give
$$
z = \frac{y -ax}{1-a} \geq \frac{a-ax}{1-a} = \frac{a(1-x)}{1-a} > 0.
$$
Since $z \in (0,1)$, it follows that $f(z)$ is strictly greater than $0$, so we have
$$
f(y) = f((1-a)z + ax) \geq (1-a)f(z) + a f(x) >af(x).
$$
Thus, we have proved (1) for $x \in (0,1)$.  For $x = 0$ or $x = 1$, we have $af(x) = 0$, but $f(y) > 0$, so (1) also holds for these endpoint cases.
A: A little geometry that follows from the definition of "concave" (it will be good to draw a picture): Let $P,Q$ be points on the graph of a concave function $f,$ and let $l(x)$ be the line through $P$ and $Q.$ Then $f\ge l$ between $P$ and $Q$ and $f\le l$ otherwise.
In your problem, let $a\in (0,1/2).$ Because $f$ is concave, the above shows the minimum of $f$ on $[a,1-a]$ must occur at one (or both) of the endpoints. Let's consider $a$ first. We want to show $f(x) < f(a)/a$ for all $x\in [0,1].$
Let $l(x)$ be the line $l$ through $(0,0)$ and $(a,f(a)).$ Then $l(x)=(f(a)/a)x.$ By concavity, $f(x)\le l(x)$ for $x\in [a,1].$ But the maximum of $l$ on $[a,1]$ is $l(1) = f(a)/a.$ And this value is never attained by $f$ since $f(1)=0.$ Thus $f(x)<f(a)/a$ for $x\in [a,1].$ 
For $x\in [0,a]$ we look at the line through $(1,0)$ and $(a,f(a)).$ The same kind of argument gives $f(x)<f(a)/(1-a).$ But note $ f(a)/(1-a) < f(a)/a.$
Conclusion: $f(x) < f(a)/a$ for all $x\in [0,1].$ Similary, $f(x) < f(1-a)/a$ for all $x\in [0,1].$ This gives the desired result.
