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Let $f:[0,1]\to\mathbb{R}$ be continuous. Suppose $f$ is differentiable on $(0,1)$ and $f(0)=0,f(1)=1$. Show that for any positive real numbers $a,b$, there are distinct points $\xi,\eta\in(0,1)$ s.t.

$$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$$

The word distinct makes this problem much harder.

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Since $a,b>0$ and $f(0)=0,f(1)=1$, $\exists \nu\in(0,1)$ s.t. $f(\nu)=\dfrac{a}{a+b}$. Hence we have $$\frac{f(\nu)-f(0)}{\nu-0}=f'(\xi)\Longrightarrow \frac{a}{a+b}=\nu f'(\xi)\Longrightarrow\frac{a}{f'(\xi)}=(a+b)\nu$$where $\xi\in(0,\nu)$.

Similarly, we have$$\frac{f(1)-f(\nu)}{1-\nu}=f'(\eta)\Longrightarrow \frac{b}{a+b}=(1-\nu) f'(\eta)\Longrightarrow\frac{b}{f'(\eta)}=(a+b)(1-\nu)$$where $\eta\in(\nu,1)$.

Therefore $$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=(a+b)\nu+(a+b)(1-\nu)=a+b$$and it's also obvious that $\xi\ne\eta$.

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