I am trying to prove a general version of Gronwall's inequality, which I saw in the book "Inequalities Involving Functions and Their Integrals and Derivatives", by Arlington M. Fink, Dragoslav Mitrinović, and Josip Pečarić in page 361. 
I don't understand where are $a$ and $b$ he talked about, and have a hard time proving this inequality as well.
The additonal condition is that $\xi$ is generated by iteration function with initial conditions as follows:
$$\xi_{0}=\lim\limits_{n\to\infty}\xi_{n}, \xi_{1}=b^{1/1-a},\xi_{n+1}=\xi_{n}-\frac{z(\xi_{n})}{z(B)-z(\xi_{n})}(B-\xi_{n}),$$ where $$B=(b\alpha)^{1/1-a}-\frac{(1-\alpha^{1/1-a})[(b\alpha)^{1/1-\alpha}-a-b(b\alpha)^{\alpha/1-\alpha}]}{(1-\alpha^{1/1-a})-b^{1+\alpha}(1-\alpha^{\alpha/1-\alpha})},$$ where $$z(\xi)=\xi-a-b\xi^{\alpha} , a=\int_{0}^{h}f(t)\text{d}t, b=c\int_{0}^{h}\left[\int_{0}^{x}\phi^{1/1-\alpha}(t)\text{d}t\right]^{1-\alpha}\text{d}x.$$
When it comes to the proof, here I give a hint:
By Holder's inequality, we can first derive $$\int_{0}^{x}\phi(x)u^{\alpha}(t)\text{d}t\leqslant \left(\int_{0}^{x}\phi^{1/1-\alpha}(t)\text{d}t\right)^{1-\alpha}\left(\int_{0}^{x}u(t)\text{d}t\right)^{\alpha},$$ then $$u(x)\leqslant f(x)+c\left(\int_{0}^{h}u(t)\text{d}t\right)^{\alpha}\left(\int_{0}^{x}\phi^{1/1-\alpha}(t)\text{d}t\right)^{1-\alpha}$$ hence $$\int_{0}^{h}u(t)\text{d}t\leqslant \int_{0}^{h}f(t)\text{d}t+c\left(\int_{0}^{h}u(t)\text{d}t\right)^{\alpha}\int_{0}^{h}\left(\int_{0}^{t}\phi^{1/1-\alpha}(s)\text{d}s\right)^{1-\alpha}\text{d}t.$$ Then by the given condition above, and consider the function $z(\xi),$ then you can derive the conclusion by the basic knowledge of limit theory in mathematical analysis.