Not a standard inequality

F(x) is positive and f(x)+f'(x) is positive. Consider, Function f(x) = e^-x But we will get f(x)+f'(x)= 0 So we consider f(x)=e^-x/k Where k --->1 (tending to 1) Then f(0)=1 And f(1) will tend to 1/e So f(0)/f(1) will be equal to "e"=2.718

From the first condition, we can assume $f(x)=e^{g(x)}$ .

From the second condition, $f(x)+f'(x)=(g'(x)+1) e^{g(x)} \ge 0 \Leftrightarrow g'(x) \ge-1$

$\implies g(1)-g(0) = \int_{0}^{1} g'(x) dx \ge \int_{0}^{1} -1 dx =-1$

So $\frac{f(0)}{f(1)}=e^{g(0)-g(1)} < e$

Note that any such $\epsilon>0$ we can find $f(x)=e^{-x}+\frac{\epsilon}{e^2}$ such that $\frac{f(0)}{f(1)}=\frac{e^2+\epsilon}{e+\epsilon} >e- \epsilon$

So $e$ is the supremum.

Letting $g(x)=e^x f(x)$ , we get $g'(x)=e^x(f(x)+f'(x))> 0\implies g$ is a strictly increasing function (from the second condition). Thus, $\frac{f(0)}{f(1)}=e\frac{g(0)}{g(1)}< e$ , for all such $f$ . Let $S$ be the set of values of all such $\frac{f(0)}{f(1)}$ , so that we have $\sup S\le e$ .

Now, lets assume that $\sup S<e$ , so that $\delta:=e-\sup S>0$ . Let us consider all functions $h$ such that $h(x)=e^{-x}+\epsilon$ , with $\epsilon> 0$ , so that $h$ satisfies both the prescribed conditions. Then, $\forall \epsilon>0$ , $\frac{h(0)}{h(1)}=\displaystyle \frac{\epsilon+1}{e^{-1}+\epsilon}\in S$ , and since $\forall \epsilon>0, \displaystyle \frac{\epsilon+1}{e^{-1}+\epsilon}>1$ , we have that $\sup S>1$ . Now, we can find $\epsilon>0$ such that $\displaystyle \frac{\epsilon+1}{e^{-1}+\epsilon}>e-\delta$ , by choosing $\epsilon$ such that $\epsilon(\sup S-1)<\delta/e$ , which is possible since $\sup S>1$ . But that means we can find an $\epsilon>0$ such that $\displaystyle \frac{\epsilon+1}{e^{-1}+\epsilon}>\sup S$ , which is a contradiction as $\displaystyle \frac{\epsilon+1}{e^{-1}+\epsilon}\in S$ by construction. Thus, we must have $\sup S=e$ $\blacksquare$

Is something wrong with my solution???.................................. I solved it by taking f(x) as y and then according to the conditions, dy/dx > -y...........Solving this results in y > exp(-x+c).........so, f(x)=exp(-x+c) + d......where c is the constant of integration and d is any really small number..........Thus putting the values, we get the minimum of the desired expression as e......