For any curve $f:\mathbb{R}\to\mathbb{R},$ the gradient of $f(x)$ at the point $x=c$ is $f'(c).$ The curve passes through the point $(c,f(c)),$ and so the equation of the tangent to the curve at $x=c$ has equation $y-f(c) = f'(c)(x-c).$ This can be rearranged to: $ g(x): (= y) = f'(c)x + (f(c) -cf'(c)). $
My question is this: Given a differentiable, concave decreasing function $f:[0,1]\to [0,1],$ with $f(0)=1$ and $f(1)=0$, does there always exist $c\in [0,1]$ such that the area of the triangle formed by $x-$ axis, the $y-$ axis, and the tangent to $y=f(x)$ at $x=c$ is less than $2\displaystyle\int_{0}^{1} f(x) dx\ ?$
Thinking about $g(0)$ and the value of $x$ s.t. $g(x) = 0,$ where $g$ is as in the first paragraph, this is equivalent to asking the following:
Given a differentiable, concave decreasing function $f:[0,1]\to [0,1],$ with $f(0)=1$ and $f(1)=0$, does there always exist $c\in [0,1]$ such that $$ \frac{ (f(c) - c f'(c)) \left( c - \frac{f(c)}{f'(c)} \right) }{2} \leq 2\int_{0}^{1} f(x) dx\ ?$$
[A counter-example would be a function $f$ such that no $c\in [0,1]$ satisfying the above property exists].
We can expand the brackets and simplify the inequality to give:
$$ 2cf(c) - c^2f'(c) - \frac{(f(c))^2}{f'(c)} \leq 4\int_{0}^{1} f(x) dx. $$
I imagine so. Maybe Holder or C-S or Minkowski are useful but I'm not really sure. I can't find any counter-examples either.
By the mean value theorem, there is a point $c$ with $f'(c)=-1$.
Then the area of the triangle can be computed as follows: on the diagonal it contains the point $(c,f(c))$. Hence the base of the triangle has length $c+f(c)$, and its total area is $\frac {(c+f(c))^2}2$. Since $c,f(c) \in [0,1]$, an upper bound of the area of the triangle is $c+f(c)$.
The integral of $f$ is at least the area of the polygon with corners $(0,0)$, $(0,1)$, $(c, f(c))$, $(1,0)$. That area is $$ c \frac{ 1+ f(c)}2 + (1-c) \frac{ f(c)}2=\frac {c+f(c)}2, $$ which is twice the upper bound of the triangle.