Let $f$ be a continuous function on $[a, b]$. Assume that there occurs a positive constant $T$ for which $$ |f(y)| \leq T \int_{a}^{y}|f(t)| d t $$ for every value $y$ in $[a, b]$. Prove that the equality $f(y)=0$ holds for any $y$ in $[a, b]$.
What could be the major strategies and original intuition behind finding a contradiction if one assumes the existence of some non-zero value? I tried to incorporate the continuity of function and indefinite integral, but it seems no success in mathematical rigor.
For $y \in [a,b]$ define
$$v(y) = e^{-T(y-a)} \int_a^y \vert f(t) \vert \ dt.$$
$v$ is differentiable as $f$ is continuous and
$$v^\prime(y) = e^{-T(y-a)} \left(-T\int_a^y \vert f(t) \vert \ dt + \vert f(y) \vert \right) \le 0$$ according to the given hypothesis. And as $v(a)=0$, you get $v(y) \le 0$ for all $y \in [a,b]$. $v$ being also obviously non-negative is the always vanishing map.
This implies that $f$ is also always vanishing as it is supposed to be continuous (the integral of a nonnegative continuous map $f$ is strictly positive as soon as $f$ takes a nonzero value).