show that $0\le \frac a{1+b}+\frac b{1+a} \le 1$ Let $0\le a < b \le 1$
prove that $0\le \frac a{1+b}+\frac b{1+a} \le 1$
This is the answer from book :
Obviously $0\le \frac a{1+b}+\frac b{1+a}$ so :
\begin{align}
&1\ge \frac a{1+b}+\frac b{1+a} \\
\iff & (1+a)(1+b) \ge a(1+a)+b(1+b)\\ 
\iff & 1-a^2\ge b^2-ab \\
\iff &(1-a)(1+a)\ge b(b-a) \tag{*}
\\ \iff & 1+a\ge b-a
\end{align}
which is obvious.  Q.E.D.
But I can't understand how we get last inequality from $(*)$.

An easier proof for me is the following:
Firstly, $\frac a{1+b}+\frac b{1+a}\geq \frac{b}{1+a}>0,\ $ as numerator and denominator of $\frac{b}{1+a}$ are both positive.
Secondly,
$$\frac a{1+b}+\frac b{1+a} = \frac{a(1+a)+b(1+b)}{(1+b)(1+a)} = \frac{a+b+a^2+b^2}{a+b+ab+1}. $$
So if we show that $ab+1 \geq a^2 + b^2, $ and then we are done. To this end:
$$\text{Since } b>a,\ \text{ and } a\geq 0,\ ab \geq a^2.\quad \text{Also, } 1\geq b, \text{ and so } 1 \geq b^2.$$
$$\text{Therefore, } ab+1 \geq a^2 + b^2.  $$
The strict inequality holds on the left side only:
$$0 < \frac a{1+b}+\frac b{1+a} \leq 1.$$
Equality on the RHS occurs when $b=1$ and $a=0.$
