I am working through Markushevich's Book 1 and I am struggling to see how to prove this inequality.
The goal is to show that for some infinitely differentiable function $f$ on a neighborhood of $x_0$, such that $f^{(k)}(x) \geq 0$ for all $k$, we have that $f(x)$ is analytic.
Following the hint, taking $x_0 - \epsilon < a < x < b < x_0 + \epsilon$, it is clear to me that $0 \leq R_n(x) \leq f(x)$, which is crucial to squeezing $R_n(x) \to 0$ after we show the following: $$ R_n(x) \leq \left(\frac{x - a}{b - a}\right)^{n + 1} R_n(b). $$
If we take the second form of the equality suggested by the hint, the resulting inequality (that I am unable to reason about) looks like $$ \int_0^1 f^{(n + 1)}((x - a)u + a)(1 - u)^n \, du \leq \frac{x - a}{b - a} \int_0^1 f^{(n + 1)}((b - a)u + a)(1 - u)^n \, du, $$ which follows from $$ \begin{aligned} R_n(x) &= \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(x - t)^n \, dt \\ &= \frac{(x - a)^n}{n!} \int_0^1 f^{(n + 1)}((x - a)u + a)(1 - u)^n \, du, \\ \end{aligned} $$ coming from $u = \dfrac{t - a}{x - a}$.
Are there any suggestions for how to proceed with proving the inequality? Trying the same $u$ substitution on both integrals is difficult because one of them uses $x - t$ and the other is $b - t$.
EDIT: I have computed the following, but the deletion of $\left(\dfrac{x - t}{b - t}\right)^n \leq 1$ doesn't feel quite right to me, but it might be correct? $$ \begin{aligned} R_n(x) &= \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(x - t)^n \, dt \\ &= \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(b - t)^n \frac{(x - t)^n}{(b - t)^n} \, dt \\ &\leq \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(b - t)^n \, dt \\ &\leq \frac{1}{n!} \int_a^b f^{(n + 1)}(t)(b - t)^n \, dt \\ \end{aligned} $$
EDIT #2: Ok, so it looks like I was on the right track. We instead use the inequality $\dfrac{x - t}{b - t} \leq \dfrac{x - a}{b - a}$ (clearing fractions and simplifying yields $(t - a)(b - x) \geq 0$ which is obviously true). Then we get $$ \begin{aligned} R_n(x) &= \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(x - t)^n \, dt \\ &= \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(b - t)^n \frac{(x - t)^n}{(b - t)^n} \, dt \\ &\leq \frac{1}{n!} \int_a^x f^{(n + 1)}(t)(b - t)^n \, dt \\ &\leq \left(\dfrac{x - a}{b - a}\right)^n \frac{1}{n!} \int_a^b f^{(n + 1)}(t)(b - t)^n \, dt \\ &= \left(\dfrac{x - a}{b - a}\right)^n R_n(b), \\ \end{aligned} $$ which seems different than what the book wanted me to prove, and could perhaps be a typo, which I originally believed to be typo'd for incorrect reasons. In retrospect, probably not a typo because there is still some funny business that was never dealt with as we fudge $x \to b$. $(b - t)^n$ remains positive and $f^{(k)}(t)$ is also positive so the value can only increase, but perhaps the extra power is the factor by which it can increase.