[不等式] 求CS.

problem:
Let $a,b,c \geq 0$ prove that:
\[ \sqrt{a(b+1)}+\sqrt{b(c+1)}+\sqrt{c(a+1)}\leq \frac{3}{2}\sqrt{(a+1)(b+1)(c+1)}\]

注意到恒等式:
\[\frac{9}{4}-\sum{\frac{a}{a+1}}\cdot \sum{\frac{1}{c+1}}=\frac{(ab+bc+ca+3abc-a-b-c-3)^2}{4(a+1)^2(b+1)^2(c+1)^2}\]
然后由Cauchy不等式证毕...

恒等式牛……

2# 天涯无际

我是这样想的。
proof:
by Cauchy-Schwar:
\[ \sqrt{a(b+1)}+\sqrt{b(c+1)}\leq \sqrt{(a+1)[(b+1)+b(c+1)]} \]
Thus,just check
\[ \sqrt{2b+bc+1}+\sqrt{c}\leq \sqrt{(b+1)(c+1)}\]
by Cauchy-Schwarz again:
\[ \sqrt{b(c+2)+1}+\sqrt{c}\leq \sqrt{[b(c+2)+1+(c+1)](1+\frac{c}{c+1})}=\sqrt{\frac{(b+1)(c+2)(2c+1)}{c+1}} \]
Or
\[ 4(c+2)(2c+1)\leq 9(c+1)^{2} \]
Using AM-GM for:
\[ 4(c+2)(2c+1)\leq 9(c+1)^{2} \]
Hence we obtain the desire result.


楼上应该是Tourish 吧。 （如果我没猜错的话。） :D

你们都好牛……

我来骗下分。本题只需要简单的A-G即可解决，我们两边同时除以$\sqrt{(a+1)(b+1)(c+1)}$
我们就可得到要证明的不等式变成：$\sqrt{\dfrac{a}{(a+1)(c+1)}}+\sqrt{\dfrac{b}{(b+1)(a+1)}}+\sqrt{\dfrac{c}{(b+1)(c+1)}}\leqslant\dfrac{3}{2}（A)$由于：$\sqrt{\dfrac{a}{(a+1)(c+1)}}\leqslant\dfrac{1}{2}(\dfrac{a}{a+1}+\dfrac{1}{c+1})(1)$
$\sqrt{\dfrac{b}{(a+1)(b+1)}}\leqslant\dfrac{1}{2}(\dfrac{b}{b+1}+\dfrac{1}{a+1})(2)$
$\sqrt{\dfrac{c}{(b+1)(c+1)}}\leqslant\dfrac{1}{2}(\dfrac{c}{c+1}+\dfrac{1}{b+1})(3)$   将(1)  (2)  (3) 叠加即可得到（A)
证毕。