Prove that $\frac{d\ f(c)-c\ f(d)}{f(d)-f(c)}=\frac{f(\xi)}{f'(\xi)}-\xi$ for some $\xi\in(c,d)$

Let $f(x)$ be a continuosly-differentiable function defined on a segment $[a,b]$ such that $a,b\in\mathbb{R}$ and $f'(x)>0$ for any $x\in(a,b)$.

Prove that for every $c,d\in\mathbb{R}$ such that $a\le c\lt d\le b$ there is a number $\xi\in(c,d)$ such that:

$$\frac{d\ f(c)-c\ f(d)}{f(d)-f(c)}=\frac{f(\xi)}{f'(\xi)}-\xi\tag{1}$$

I have started from the basic fact that there is some $\xi\in(c,d)$ such that:

$$\frac{f(d)-f(c)}{d-c}=f'(\xi)$$

That was just a start that led me nowhere. I have also tried to rewite (1) in a slightly different way:

$$\frac{d\ f(c)-c\ f(d)}{f(d)-f(c)}=\frac{\frac{f(c)}c-\frac{f(d)}d}{{\frac1c}\frac{f(d)}d-{\frac1d}\frac{f(c)}c}$$

...and then introduce function $g(x)=\frac{f(x)}x$ and play with it. And that was a dead end too.

Any hint or help will be much appreciated.

Solution 1:

According to Cauchy's version of Mean Value Theorem, there exists an $\xi\in(c,d)$, such that $$ \frac{cf(d)-df(c)}{f(d)-f(c)}=\frac{\frac{c}{f(c)}-\frac{d}{f(d)}}{\frac{1}{f(c)}-\frac{1}{f(d)}} =\frac{\left(\frac{\xi}{f(\xi)}\right)'}{\left(\frac{1}{f(\xi)}\right)'} =\frac{\frac{f(\xi)-\xi f'(\xi)}{f^2(\xi)}}{-\frac{f'(\xi)}{f^2(\xi)} }=-\frac{f(\xi)-\xi f'(\xi)}{f'(\xi)}=\xi-\frac{f(\xi)}{f'(\xi)} $$

Solution 2:

Let $\kappa=\frac{df(c)-cf(d)}{f(d)-f(c)}$, and see $$\int_c^d\frac{1}{\kappa+\xi}d\xi=\ln\left(\frac{\kappa+d}{\kappa+c}\right)=\ln\left(\frac{\frac{(d-c)f(d)}{f(d)-f(c)}}{\frac{(d-c)f(c)}{f(d)-f(c)}}\right)=\ln(f(d)/f(c)).$$ Also, $$\int_c^d\frac{f'(\xi)}{f(\xi)}d\xi=\ln f(\xi)\bigg|_c^d=\ln(f(d)/f(c)).$$ So, since these are both continuous functions of $\xi$, with the same integral, they must agree somewhere.