Find all continuous function $f:[0,1]\rightarrow\mathbb{R}$ which is differentiable on $(0,1)$ and $$f(0)=f(1)=\frac{2021}{2020}\textrm{ while }2019f'(x)+2020f(x)\geq2021,\forall x\in(0,1).$$
The Instructor hints me to use Rolle's Theorem, but all I can get is $$\exists x_0\in(0,1):f'(x_0)=0\Rightarrow f(x_0)\geq\frac{2021}{2020}.$$ I guess the only such function is $f\equiv2021/2020.$
Any hints/ideas/comments are appericated. Thank you!
Let $$F(x)=\exp\left(\frac{2020}{2019}x\right)\cdot\left(f(x)-\frac{2021}{2020}\right),\qquad x\in [0,1].$$ Then $F$ is differentiable on $(0,1)$ with $$F'(x)=\frac1{2019}\exp\left(\frac{2020}{2019}x\right)\cdot\left(2020f(x)-2021+2019f'(x)\right)\geq 0.$$ Hence $F$ is nondecreasing. Also we have $F(0)=F(1)=0$, so $F(x)=0$ for all $x\in[0,1]$. Therefore, $$f(x)=\frac{2021}{2020},\qquad\forall x\in[0,1].$$
Remark. If $F$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with $F(0)=F(1)$ and $F'(x)\geq 0$ for all $x\in (0,1)$, then $F$ is constant on $[0,1]$. This is a consequence of the mean value theorem, which can be proved using Rolle's theorem, see the proof on Wikipedia.