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I want to prove that $x<\frac{2x}{2-x}, \forall x \in (0,1)$, by using the mean value theorem.

So, consider $f(x)=\frac{2x}{2-x} -x$. $f(0)=0$. $f´(x)=\frac{2x-2}{(2-x)^2} - 1$ and $f'(x)<0, \forall x \in (0,1)$. By the mean value theorem:

$$\exists c \in (0,1)~~~\text{such that}~~~f(x)-f(0) = f'(c)(x-0)~~~\rightarrow~~~f(x)<0 ~~~\rightarrow~~~x> \frac{2x}{2-x}$$

So, it doesn't work.

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It turns out that $$f'(x)=\frac{4}{(2-x)^2}-1.$$ This is safely positive on our interval, so the MVT argument goes through.

Remark: It seems you were asked to use MVT explicitly. The ordinary way is to use the theorem, derived using the MVT, that if the derivative of a function is positive on an interval, then the function is increasing on that interval.

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