Let $\mathbb{D}$ be the open unit disc in the complex plane. Let $\gamma: [a,b] \longrightarrow\mathbb{D}$ be a piecewise smooth curve such that $\gamma(a)=0$ and $\gamma(b)=r\in (0,1)$ . Let $\gamma_r$ be the real part of $\gamma$ and let $\gamma_{r+}$ be $max\{\gamma_r, 0\}$. My question is
1) are $\gamma_r$ and $\gamma_{r+}$ also piecewise smooth?
2) How is the expression below true? $\int_{a}^{b}\frac{|\gamma '(t)|}{1-{|\gamma (t)|}^2}dt\geq \int_{a}^{b}\frac{|\gamma_{r} '(t)|}{1-{|\gamma_{r}(t)|}^2}dt\geq \int_{a}^{b}\frac{|\gamma_{r+} '(t)|}{1-{|\gamma_{r+}(t)|}^2}dt$
1) is trivial by definition of piecewise smooth - a complex function is piecewise smooth iff the real and imaginary parts are so, and obviously taking the maximum of two piecewise smooth functions, you get another such which is intuitively clear, but also can be seen from the expression $2\max${$a,b$}$=a+b+|a-b|$
2) the function $\frac{x}{1-x}=\frac{1}{1-x}-1$ is strictly increasing on $[0,1)$ and obviously $1>|\gamma '(t)| \ge |\gamma_r '(t)|\ge |\gamma_{r+} '(t)| \ge 0$, hence the integrands decrease term by term in the corresponding integrals for each $t$ where they are smooth, while the finitely many $t$ for which they are not smooth make no diiference when integrating