来自西西解疑群中的一个不等式

陈洪葛 posted @ Aug 25, 2013 11:50:28 AM in 不等式 , 1044 阅读

 Problem

设$a,b,c \in \mathbf{R}$.且$ab+bc+ca=11$.求
\[ \sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91} \]
的最小值。


Solution

Cauchy-Schwarz

\[ [(a^2+1)+20](5+20)\geq \left[\sqrt{5(a^2+1)}+20\right]^{2}   \]
 \[      [ 2(b^2+1)+12](4+12)\geq \left[\sqrt{8(b^2+1)}+12\right]^2   \]
 \[      [(c^2+1)+90](10+90)\geq \left[\sqrt{10(c^2+1)}+90\right]^2 \]
得到
\[ \sqrt{a^2+21}\geq \sqrt{\frac{1}{5}\left(a^2+1\right)}+4 \]
\[ \sqrt{2b^2+14}\geq \sqrt{\frac{1}{2}\left(b^2+1\right)}+3 \]
\[ \sqrt{c^2+91} \geq \sqrt{\frac{1}{10}\left(c^2+1\right)}+9 \]
\begin{align*}
\sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}&\geq 16 +\sqrt{\frac{1}{5}\left(a^2+1\right)}+\sqrt{\frac{1}{2}\left(b^2+1\right)}+\sqrt{\frac{1}{10}\left(c^2+1\right)}\\
&\geq 16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}
\end{align*}
注意到恒等式
\[ (a^2+1)(b^2+1)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2\geq (ab+bc+ca-1)^2\ge 100 \]

\[ 16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}\geq 19 \]
\[ \sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}\geq 19 \]
等号成立当$a=2,b=1,c=3 $                                                        $\square$
 


登录 *


loading captcha image...
(输入验证码)
or Ctrl+Enter