[不等式] kuing来CS吧

Let $a,b,c$ be positive real numbers. Prove that :
\[ \frac{bc}{3a^2+b^2+c^2}+\frac{ca}{3b^2+c^2+a^2}+\frac{ab}{3c^2+a^2+b^2}\le\frac{3}{5}. \]
(Vasc and Pham Kim Hung, 2005)

硬开后等价为
\[ 9\sum{a^6}-50\sum{a^3b^3}-15\sum{a^5(b+c)}+39\sum{a^4(b^2+c^2)}-5abc\sum{a^3}-20abc\sum{a^2(b+c)}+114a^2b^2c^2\geq 0 \]

it is not easy to give a CS proof, but we can give another proof easily.

it is not easy to give a CS proof, but we can give another proof easily.
zdyzhj 发表于 2013-1-26 21:41

牛啊

3# zdyzhj


true!
Let $ x=\min{(x,y,z)}$ it can be write
\[ (x^2+2y^2+2z^2)(3x-y-z)^2(x-y)(x-z)+[14x^4-19x^3(y+z)+25x^2(y^2+z^2)-17x^2yz-13x(y+z)(y^2+z^2)+9(y^4+z^4)+yz(y^2+z^2)+28y^2z^2](y-z)^2\geq 0\]

3# zdyzhj


true!
Let $ x=\min{(x,y,z)}$ it can be write
\[ (x^2+2y^2+2z^2)(3x-y-z)^2(x-y)(x-z)+[14x^4-19x^3(y+z)+25x^2(y^2+z^2)-17x^2yz-13x(y+z)(y^2+z^2)+9(y^4+z^4)+yz(y^2+z^2)+28y^2z^2](y-z) ...
pxchg1200 发表于 2013-1-26 22:36

机器？

5# pxchg1200    6# yes94

好久没见过这种分拆了，当年看褚小光他们经常用，我倒用得很少（对上一次用应该是在这里，灰常bao力），也不太熟悉中间的配凑技巧，反正不用软件辅助我是不敢玩这种分拆的……
早前看过褚小光写的关于这种分拆的文章，一时找不到在哪了……

7# kuing


Well,Equality occurs when $ a=b=c$ or $a=2,b=c=3$ up to permutation.with the two euqality cases,I wonder there exist a excellent CS proof or not...