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Let $f:[0,a]\rightarrow[0,\infty)$ be a continuous function such that $$f(t)\leq e^{\int_{0}^{t}f(s)ds}-1$$ for all $t\in[0,a]$. Prove that $f\equiv0$.

I have thought like this: Assume $F(t)=\int_{0}^{t}f(s)ds\implies F'(t)=f(t)$ . Then $F'(t)\leq e^{F(t)}-1$. Now how can proceed.

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Since $$ F'(t) \leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality $$e^{-F(t)-t} (F'(t)+1) \leq e^{-t}$$ $$ \implies \frac {d}{dt} e^{-F(t)-t} \geq \frac {d}{dt} e^{-t}$$ $$\implies e^{-F(t)-t} \geq e^{-t} $$ $$ \implies e^{F(t)} \leq 1$$ $$\implies F(t)\leq 0$$ But we know $F$ is non negative and hence $F\equiv 0$ So $$ f\equiv 0$$

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