[不等式] just CS it,(1)

1.Let $a,b,c$ be real numbers.prove that:
\[ (a^{2}+3)(b^{2}+3)(c^{2}+3)\geq 4(a+b+c+1)^{2}\]


2.for postive numbers $ x_{1},x_{2},\cdots,x_{n}$ we have
\[ x_{1}+x_{2}+\cdots+x_{n}=n \]
prove that:
\[ \frac{1}{x_{1}^{2}-x_{1}+n}+\frac{1}{x_{2}^{2}-x_{2}+n}+\cdots+\frac{1}{x_{n}^{2}-x_{n}+n}\leq 1 \]

Have fun!

咦？ 这两个应该不难的啊，怎么没人做？！

Solution 1 :
WLOG we can assume that : $ (a^{2}-1)(b^{2}-1)\geq 0 $
Therefore,$ (a^{2}+3)(b^{2}+3)=(a^{2}-1)(b^{2}-1)+4(a^{2}+b^{2}+2) $
by CS
$ (a^{2}+b^{2}+1+1)(1+1+c^{2}+1)\geq (a+b+c+1)^{2} $
Done!

Solution 2:
by CS
\[ (x_{1}^{2}+(n-x_{1}))(1+(n-x_{1}))\geq (x_{1}+(n-x_{1}))^{2}=n^{2} \]
Therefore:
\[ \frac{1}{x_{1}^{2}-x_{1}+n}\leq\frac{n+1-x_{1}}{n^{2}} \]
sum it up,the result follows