回復 1# tsyr 的帖子
\(\begin{align}
& \sqrt{{{a}^{3}}+1}=\sqrt{\left( a+1 \right)\left( {{a}^{2}}-a+1 \right)}\le \frac{a+1+{{a}^{2}}-a+1}{2}=\frac{{{a}^{2}}+2}{2} \\
& \sqrt{\left( {{a}^{3}}+1 \right)\left( {{b}^{3}}+1 \right)}\le \frac{{{a}^{2}}+2}{2}\times \frac{{{b}^{2}}+2}{2}=\frac{\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)}{4} \\
& \frac{{{a}^{2}}}{\sqrt{\left( {{a}^{3}}+1 \right)\left( {{b}^{3}}+1 \right)}}\ge \frac{{{a}^{2}}}{\frac{\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)}{4}}=\frac{4{{a}^{2}}}{\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)} \\
\end{align}\)
接下來只要證明\(\frac{{{a}^{2}}}{\left( {{a}^{2}}+2 \right)\left( {{b}^{2}}+2 \right)}+\frac{{{b}^{2}}}{\left( {{b}^{2}}+2 \right)\left( {{c}^{2}}+2 \right)}+\frac{{{c}^{2}}}{\left( {{c}^{2}}+2 \right)\left( {{a}^{2}}+2 \right)}\ge \frac{1}{3}\)即可
令\(x={{a}^{2}},y={{b}^{2}},z={{c}^{2}}\)
可得
\(\begin{align}
& xy+yz+zx\ge 3\sqrt[3]{{{\left( xyz \right)}^{2}}}=48 \\
& x+y+z\ge 3\sqrt[3]{xyz}=12 \\
\end{align}\)
即證明
\(\begin{align}
& \frac{x}{\left( x+2 \right)\left( y+2 \right)}+\frac{y}{\left( y+2 \right)\left( z+2 \right)}+\frac{z}{\left( z+2 \right)\left( x+2 \right)}\ge \frac{1}{3} \\
& \Leftrightarrow 3\left[ x\left( z+2 \right)+y\left( x+2 \right)+z\left( y+2 \right) \right]\ge \left( x+2 \right)\left( y+2 \right)\left( z+2 \right) \\
& \Leftrightarrow 3\left( xy+yz+zx \right)+6\left( x+y+z \right)\ge xyz+2\left( xy+yz+zx \right)+4\left( x+y+z \right)+8 \\
& \Leftrightarrow xy+yz+zx+2\left( x+y+z \right)\ge 64+8=72 \\
\end{align}\)
[ 本帖最後由 thepiano 於 2015-2-28 09:54 PM 編輯 ]
