<s>[INST] 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 $(*)$.
 [/INST] I guess this is a typo, they meant to say
$$ 1-a \ge b-a,$$
which is true from
$$ (1-a)(1+a) \ge b(b-a)$$
since $1+a \ge 1 \ge b$.
(Of course, $1-a\ge b-a$ is the same as $1\ge b$)
</s>