# 1.
Assume $a+b+c = 1$, then the inequality becomes
$$\sum_{cyc}\sqrt{\frac{a}{b+1}}\leq \frac{3}{2}$$

By Cauchy-Schwarz,
$$
\sum_{cyc}\sqrt{\frac{a}{b+1}} \leq \sqrt{\sum_{cyc}(a+1)} \sqrt{\sum_{cyc}\frac{a}{(b+1)(a+1)}} = 2\sqrt{\sum_{cyc}\frac{a}{(b+1)(a+1)}}
$$

Let $s_2 = ab + bc + ac$, $s_3 = abc$, we have

$$
\sum_{cyc}\frac{a}{(b+1)(a+1)} = \sum_{cyc}\frac{a(c+1)}{(a+1)(b+1)(c+1)}=\frac{1+s_2}{2+s_2+s_3}
$$
By Schur's inequality(小蓝本不等式 pp.19, 变形II)
$$
s_3 \geq \frac{4s_2 - 1}{9}
$$
Plug it into the third line formula:
$$
\sum_{cyc}\frac{a}{(b+1)(a+1)} \leq \frac{1+s_2}{2+s_2+\frac{4s_2-1}{9}} = \frac{9+9s_2}{17+13s_2}
$$
To show $\sum_{cyc}\sqrt{\frac{a}{b+1}}\leq \frac{3}{2}$, it suffices to show
$$
\frac{9+9s_2}{17+13s_2} \leq \frac{9}{16}
$$
And it's simply true as $s_2\leq 1/3$.

# 2.

I underestimated it. It's hrad. I list some references to it:
1. https://artofproblemsolving.com/community/c4h2026588p14265546
2. https://artofproblemsolving.com/community/c6h2035758p14374093
3. https://artofproblemsolving.com/community/c6h176214p973283
4. https://artofproblemsolving.com/community/c6h205602p1131751
5. https://artofproblemsolving.com/community/c6h2027246p14272357