一个有意思的不等式

陈洪葛 posted @ Jan 30, 2013 10:57:29 AM in 不等式 , 1166 阅读

Let $a,b,c>0$ with $abc=1$ Prove that:
\[ \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \]
Proof:
We have
\[ \frac{2}{a+2}-\frac{b}{ab+b+1}-\frac{1}{a+b+1}=\frac{a(b-1)^2}{(a+2)(ab+b+1)(a+b+1)}\geq0 \]
we get
\[ 2\sum{\frac{1}{a+2}}\geq \sum{\frac{b}{ab+b+1}}+\sum{\frac{1}{a+b+1}}\]
since
\[ \sum{\frac{b}{ab+b+1}}=1 \]
This inequality yields
\[ \sum{\frac{1}{a+2}}-\sum{\frac{1}{a+b+1}}\geq 1-\frac{1}{a+2} \]
Thus,it's suffice to check
\[ \frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\geq 1 \]
Which is obvious from AM-GM
\[ \frac{a}{a+2}=\frac{a}{a+2(abc)^{\frac{1}{3}}}\geq \frac{a^{\frac{2}{3}}}{a^{\frac{2}{3}}+b^{\frac{2}{3}}+c^{\frac{2}{3}}} \]
Hence we are done!


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