由人教群的一道题即兴编题

kuing

编辑-数理化解t (6467****)  21:08:57
求证:根号下x/y+z+根号下y/x+z+根号下z/x+y大于2
谁帮我？
Admin-kuing  21:38:04
我来编道题：
“ 根号下x/y+z+根号下y/x+z+根号下z/x+y大于2 ”
这个表达总共能够理解成多少种不等式？
例如：
\begin{gather*}
\sqrt{x}/y+z+\sqrt{y}/x+z+\sqrt{z}/x+y>2 \\
\sqrt{x/y}+z+\sqrt{y/x}+z+\sqrt{z/x}+y>2 \\
\sqrt{x/y+z+\sqrt{y/x+z+\sqrt{z/x+y}}}>2 \\
\cdots
\end{gather*}
等……

tommywong

a=1,2,3,4,5,6,7,8,9
b=4,5,6,7,8,9
c=7,8,9
b=5时,a≠4
b=6时,a≠4,5
b=7时,a≠4,5,6
b=8时,a≠4,5,6,7
b=9时,a≠4,5,6,7,8
c=8时,a,b≠7
c=9时,a,b≠7,8
似乎很混乱......
[(4,5,?)+(4,6,?)+(5,6,?)+(4,7,?)+(5,7,?)+(6,7,?)+(4,8,?)+(5,8,?)+(6,8,?)+(7,8,?)+(4,9,?)+(5,9,?)+(6,9,?)+(7,9,?)+(8,9,?)]∪
[(7,?,8)∪(?,7,8)∪(7,?,9)∪(?,7,9)∪(8,?,9)∪(?,8,9)]

=[(4,5,?)+(4,6,?)+(5,6,?)+(4,7,?)+(5,7,?)+(6,7,?)+(4,8,?)+(5,8,?)+(6,8,?)+(7,8,?)+(4,9,?)+(5,9,?)+(6,9,?)+(7,9,?)+(8,9,?)]∪
[(7,?,8)+(?,7,8)+(7,?,9)+(?,7,9)+(8,?,9)+(?,8,9)-(7,7,8)-(7,7,9)-(7,8,9)-(8,7,9)-(8,8,9)]

=(4,5,?)+(4,6,?)+(5,6,?)+(4,7,?)+(5,7,?)+(6,7,?)+(4,8,?)+(5,8,?)+(6,8,?)+(7,8,?)+(4,9,?)+(5,9,?)+(6,9,?)+(7,9,?)+(8,9,?)
+(7,?,8)+(?,7,8)+(7,?,9)+(?,7,9)+(8,?,9)+(?,8,9)-(7,7,8)-(7,7,9)-(7,8,9)-(8,7,9)-(8,8,9)
-(7,8,8)-(7,9,8)-(4,7,8)-(5,7,8)-(6,7,8)-(7,8,9)-(7,9,9)-(4,7,9)-(5,7,9)-(6,7,9)-(8,8,9)-(4,8,9)-(5,8,9)-(6,8,9)-(7,8,9)+(7,8,9)=71

3*6*9-71=91

其妙

回复 2# tommywong
计数大神呀！

tommywong

ゆ？

$\begin{pmatrix}
X & X & X & X & X & X & X & X & X\\
0 & 0 & 0 & X & X & X & X & X & X\\
0 & 0 & 0 & 0 & 0 & 0 & X & X & X
\end{pmatrix}
=\begin{pmatrix}
X & X & X & X & X & X & X & X & X\\
0 & 0 & 0 & X & X & X & X & X & X\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0
\end{pmatrix}
+\begin{pmatrix}
X & X & X & X & X & X & 0 & X & X\\
0 & 0 & 0 & X & X & X & 0 & X & X\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
\end{pmatrix}
+\begin{pmatrix}
X & X & X & X & X & X & 0 & 0 & X\\
0 & 0 & 0 & X & X & X & 0 & 0 & X\\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}$

$\displaystyle \sum_{i=0}^{c-1}\sum_{j=0}^{b-1-i} a-i-j=C_a^1 C_b^1 C_c^1 -C_b^2 C_c^1-C_a^1 C_c^2 +C_c^3$

$C_9^1 C_6^1 C_3^1 -C_6^2 C_3^1-C_9^1 C_3^2 +C_3^3=91$

tommywong

$a=C_a^1$

$\displaystyle \sum_{i=0}^{b-1} a-i=(C_a^1)C_b^1-C_b^2$

$\displaystyle \sum_{j=0}^{c-1} \sum_{i=0}^{b-j-1} a-i-j=((C_a^1)C_b^1-C_b^2)C_c^1-(C_a^1)C_c^2+C_c^3$

$\displaystyle \sum_{k=0}^{d-1} \sum_{j=0}^{c-k-1} \sum_{i=0}^{b-j-k-1} a-i-j-k=(((C_a^1)C_b^1-C_b^2)C_c^1-(C_a^1)C_c^2+C_c^3)C_d^1-((C_a^1)C_b^1-C_b^2)C_d^2+(C_a^1)C_d^3-C_d^4$

把a=b=c=d=n代进去，那东西居然出现了

$\displaystyle n,\frac{1}{2}n^2+\frac{1}{2}n,\frac{1}{6}n^3+\frac{1}{2}n^2+\frac{1}{3}n,\frac{1}{24}n^4+\frac{1}{4}n^3+\frac{11}{24}n^2+\frac{1}{4}n$