<s>[INST] Prove that $\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9}$. Let $a,b,c$ be non-negative real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$. Prove that 
$\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9}$.
Here equality holds when $a=b=c=1$ so by using AM-GM on $c+1+1$ we get 
$\sqrt{(a+b+1)(c+2)}+\sqrt{(b+c+1)(a+2)}+\sqrt{(c+a+1)(b+2)}\ge{9(abc)^{1/6}}$ but this is not right choice because by am-gm on given condition we have $(abc)^{1/6}\le{1}$. I tried Cauchy-Schwartz, too.
 [/INST] Cauchy Schwarz actually works:
$$
\sum_{cyc}\sqrt{(a+b+1)(1+1+c)}≥\sum_{cyc}\sqrt{\left(\sqrt a+\sqrt b+\sqrt c\right)^2}=3(\sqrt a+\sqrt b+\sqrt c)=9
$$
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