回復 4# 王重鈞 的帖子
我是算幾炸到底,一路黑到底,慘不忍睹
以下使用符號 \( \sum\limits _{cyclic}f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b) \)
\( 2a+b+c=(a+b)+(a+c)\geq2\sqrt{(a+b)(a+c)} \) (算幾不等式)
故 \( \frac{1}{(2a+b+c)^{2}}\leq\frac{1}{4}\cdot\frac{1}{(a+b)(a+c)} \),同理有 \( \frac{1}{(2b+c+a)^{2}}\leq\frac{1}{4}\cdot\frac{1}{(b+c)(b+a)} \), \( \frac{1}{(2c+a+b)^{2}}\leq\frac{1}{4}\cdot\frac{1}{(c+a)(c+b)} \)
因此 \( \frac{1}{(2a+b+c)^{2}}+\frac{1}{(2b+c+a)^{2}}+\frac{1}{(2c+a+b)^{2}}\leq\frac{1}{4}\left[\frac{1}{(a+b)(a+c)}+\frac{1}{(b+c)(b+a)}+\frac{1}{(c+a)(c+b)}\right]=\frac{1}{2}\frac{a+b+c}{(a+b)(b+c)(c+a)} \)....(1)
由 \( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c \) 可得 \( \frac{ab+bc+ca}{abc}=a+b+c \Rightarrow\frac{abc(a+b+c)}{ab+bc+ca} = 1 \)...(2)
(1)(2) \( \Rightarrow\frac{1}{(2a+b+c)^{2}}+\frac{1}{(2b+c+a)^{2}}+\frac{1}{(2c+a+b)^{2}}\leq\frac{1}{2}\frac{a+b+c}{(a+b)(b+c)(c+a)}\cdot\frac{abc(a+b+c)}{ab+bc+ca}=\frac{1}{2}\frac{\sum\limits _{cyclic}(a^{3}bc+2ab^{2}c^{2})}{\sum\limits _{cyclic}(a^{3}b^{2}+a^{3}c^{2}+2a^{3}bc+4ab^{2}c^{2})} \)...(3)
算幾不等式有 \( a^{3}b^{2}+a^{3}c^{2}\geq2a^{3}bc \Rightarrow\sum\limits _{cyclic}\left(a^{3}b^{2}+a^{3}c^{2}\right)\geq\sum\limits _{cyclic}2a^{3}bc \)...(4)
算幾不等式有 \( a^{3}b^{2}+ab^{2}c^{2}\geq2a^{2}b^{2}c, a^{3}c^{2}+ab^{2}c^{2}\geq2a^{2}bc^{2} \)
故 \( \sum\limits _{cyclic}(a^{3}b^{2}+a^{3}c^{2}+2ab^{2}c^{2})\geq\sum\limits _{cyclic}(2a^{2}b^{2}c+2a^{2}bc^{2}) \)
\( \Rightarrow\sum\limits _{cyclic}(a^{3}b^{2}+a^{3}c^{2}+2ab^{2}c^{2})\geq\sum\limits _{cyclic}4ab^{2}c^{2} \)
\( \Rightarrow\sum\limits _{cyclic}(a^{3}b^{2}+a^{3}c^{2})\geq\sum\limits _{cyclic}2ab^{2}c^{2} \)...(5)
(4)(5) \( \Rightarrow\frac{1}{3}\sum\limits _{cyclic}\left(a^{3}b^{2}+a^{3}c^{2}\right)+\frac{2}{3}\sum\limits _{cyclic}\left(a^{3}b^{2}+a^{3}c^{2}\right)\geq\sum\limits _{cyclic}\left(\frac{2}{3}a^{3}bc+\frac{4}{3}ab^{2}c^{2}\right) \)...(6)
(3)(6) \( \Rightarrow\frac{1}{(2a+b+c)^{2}}+\frac{1}{(2b+c+a)^{2}}+\frac{1}{(2c+a+b)^{2}}\leq\frac{1}{2}\frac{\sum\limits _{cyclic}(a^{3}bc+2ab^{2}c^{2})}{\sum\limits _{cyclic}(\frac{8}{3}a^{3}bc+\frac{16}{3}ab^{2}c^{2})}=\frac{3}{16} \).
而當 \( a=b=c=1 \) 時 \( \frac{1}{(2a+b+c)^{2}}+\frac{1}{(2b+c+a)^{2}}+\frac{1}{(2c+a+b)^{2}}=\frac{3}{16} \),故所求最大值為 \( \frac{3}{16} \).