那么下面,就可以一起看看这题的详细过程了(当然,没事干 有兴趣的读者也可以动笔尝试一下~):
x , y , z > 0 , 求证: ( x y + y z + z x ) ⋅ [ 1 ( x + y ) 2 + 1 ( y + z ) 2 + 1 ( z + x ) 2 ] ≥ 9 4 . x,y,z>0,\text{求证:}\left( xy+yz+zx \right) \cdot \left[ \frac{1}{\left( x+y \right) ^2}+\frac{1}{\left( y+z \right) ^2}+\frac{1}{\left( z+x \right) ^2} \right] \ge \frac{9}{4}. x,y,z>0,求证:(xy+yz+zx)⋅[(x+y)21+(y+z)21+(z+x)21]≥49.
证明:经过尝试可以知道,这题对于放缩有着极高的要求,因此在技巧性过强的时候,可以考虑使用暴力展开这一方法。
原 ⇔ 4 ( x y + y z + z x ) ⋅ ∑ [ ( x + y ) 2 ⋅ ( y + z ) 2 ] ≥ 9 ⋅ ( ∑ x 2 y + ∑ x y 2 + 2 x y z ) 2 而 ∑ [ ( x + y ) ⋅ ( y + z ) ] 2 = ∑ ( x y + y z + z x + y 2 ) 2 = ∑ [ ( x y + y z + z x ) 2 + y 4 + 2 y 2 ⋅ ( x y + y z + z x ) ] = ∑ ( ∑ x 2 y + 2 ∑ x 2 y z + y 4 + 2 y 2 ∑ x y ) = 3 ∑ x 2 y 2 + 6 ∑ x 2 y z + ∑ x 4 + 2 ⋅ ( ∑ x y ) ⋅ ( ∑ x 2 ) . \text{原}\Leftrightarrow 4\left( xy+yz+zx \right) \cdot \sum{\left[ \left( x+y \right) ^2\cdot \left( y+z \right) ^2 \right]}\ge 9\cdot \left( \sum{x^2y+\sum{xy^2}+2xyz} \right) ^2\\\text{而}\sum{\left[ \left( x+y \right) \cdot \left( y+z \right) \right] ^2=\sum{\left( xy+yz+zx+y^2 \right) ^2}}\\=\sum{\left[ \left( xy+yz+zx \right) ^2+y^4+2y^2\cdot \left( xy+yz+zx \right) \right]}\\=\sum{\left( \sum{x^2y}+2\sum{x^2yz+y^4+2y^2\sum{xy}} \right)}\\=3\sum{x^2y^2+6\sum{x^2yz+\sum{x^4}+2\cdot \left( \sum{xy} \right) \cdot \left( \sum{x^2} \right) .}} 原⇔4(xy+yz+zx)⋅∑[(x+y)2⋅(y+z)2]≥9⋅(∑x2y+∑xy2+2xyz)2而∑[(x+y)⋅(y+z)]2=∑(xy+yz+zx+y2)2=∑[(xy+yz+zx)2+y4+2y2⋅(xy+yz+zx)]=∑(∑x2y+2∑x2yz+y4+2y2∑xy)=3∑x2y2+6∑x2yz+∑x4+2⋅(∑xy)⋅(∑x2).
左式 4 = ( x y + y z + z x ) ⋅ ( 3 ∑ x 2 y 2 + 8 ∑ x 2 y z + ∑ x 4 + 2 ∑ x 3 y + 2 ∑ x y 3 ) = 3 ⋅ ( ∑ x y ) ⋅ ( ∑ x 2 y 2 ) + 8 ⋅ ( ∑ x y ) ⋅ ( ∑ x 2 y z ) + ( ∑ x y ) ⋅ ( ∑ x 4 ) + 2 ⋅ ( ∑ x y ) ⋅ ( ∑ x 3 y ) + 2 ⋅ ( ∑ x y ) ⋅ ( ∑ x y 3 ) = 3 ⋅ ( ∑ x 3 y 3 + ∑ x 2 y 3 z + ∑ x 3 y 2 z ) + 8 ⋅ ( ∑ x 2 y 3 z + ∑ x 2 y 2 z 2 + ∑ x 3 y 2 z ) + ( ∑ x 5 y + ∑ x 4 y z + ∑ x 5 z ) + 2 ⋅ ( ∑ x 4 y 2 + ∑ x 3 y 2 z + ∑ x 4 y z ) + 2 ⋅ ( ∑ x 2 y 4 + ∑ x y 4 z + ∑ x 2 y 3 z ) = 3 ∑ x 3 y 3 + 13 ∑ x 2 y 3 z + 13 ∑ x 3 y 2 z + 24 x 2 y 2 z 2 + ( ∑ x 5 y + ∑ x 5 z ) + 5 ∑ x 4 y z + 2 ⋅ ( ∑ x 4 y 2 + ∑ x 2 y 4 ) \frac{\text{左式}}{4}=\left( xy+yz+zx \right) \cdot \left( 3\sum{x^2y^2+8\sum{x^2yz}+\sum{x^4}+2\sum{x^3y}+2\sum{xy^3}} \right) \\=3\cdot \left( \sum{xy} \right) \cdot \left( \sum{x^2y^2} \right) +8\cdot \left( \sum{xy} \right) \cdot \left( \sum{x^2yz} \right) +\left( \sum{xy} \right) \cdot \left( \sum{x^4} \right) +2\cdot \left( \sum{xy} \right) \\\,\, \cdot \left( \sum{x^3y} \right) +2\cdot \left( \sum{xy} \right) \cdot \left( \sum{xy^3} \right) \\=3\cdot \left( \sum{x^3y^3+\sum{x^2y^3z+\sum{x^3y^2z}}} \right) +8\cdot \left( \sum{x^2y^3z+\sum{x^2y^2z^2}+\sum{x^3y^2z}} \right) \\\,\, +\left( \sum{x^5y}+\sum{x^4yz}+\sum{x^5z} \right) +2\cdot \left( \sum{x^4y^2}+\sum{x^3y^2z}+\sum{x^4yz} \right) \\\,\, +2\cdot \left( \begin{array}{c} \sum{\begin{array}{c} \begin{array}{c} x^2y^4\\\end{array}\\\end{array}}+\sum{xy^4z}+\sum{x^2y^3z}\\\end{array} \right) \\=3\sum{x^3y^3}+13\sum{x^2y^3z}+13\sum{x^3y^2z}+24x^2y^2z^2+\left( \sum{x^5y}+\sum{x^5z} \right) \\\,\, +5\sum{x^4yz}+2\cdot \left( \sum{x^4y^2}+\sum{x^2y^4} \right) 4左式=(xy+yz+zx)⋅(3∑x2y2+8∑x2yz+∑x4+2∑x3y+2∑xy3)=3⋅(∑xy)⋅(∑x2y2)+8⋅(∑xy)⋅(∑x2yz)+(∑xy)⋅(∑x4)+2⋅(∑xy)⋅(∑x3y)+2⋅(∑xy)⋅(∑xy3)=3⋅(∑x3y3+∑x2y3z+∑x3y2z)+8⋅(∑x2y3z+∑x2y2z2+∑x3y2z)+(∑x5y+∑x4yz+∑x5z)+2⋅(∑x4y2+∑x3y2z+∑x4yz)+2⋅(∑x2y4+∑xy4z+∑x2y3z)=3∑x3y3+13∑x2y3z+13∑x3y2z+24x2y2z2+(∑x5y+∑x5z)+5∑x4yz+2⋅(∑x4y2+∑x2y4)
右式 9 = ( ∑ x 2 y + ∑ x y 2 + 2 x y z ) 2 = ( ∑ x 2 y ) 2 + ( ∑ x y 2 ) 2 + 4 x 2 y 2 z 2 + 2 ⋅ ( ∑ x 2 y ⋅ ∑ x y 2 ) + 4 x y z ⋅ ∑ x 2 y + 4 x y z ⋅ ∑ x y 2 = ( ∑ x 4 y 2 + 2 ∑ x 2 y 3 z ) + ( ∑ x 2 y 4 + 2 ∑ x y 3 z 2 ) + 4 x 2 y 2 z 2 + 2 ⋅ ( ∑ x 3 y 3 + ∑ x 2 y 2 z 2 + ∑ x 4 y z ) + 4 x y z ⋅ ∑ x 2 y + 4 x y z ⋅ ∑ x y 2 \frac{\text{右式}}{9}=\left( \sum{x^2y}+\sum{xy^2}+2xyz \right) ^2\\=\left( \sum{x^2y} \right) ^2+\left( \sum{xy^2} \right) ^2+4x^2y^2z^2+2\cdot \left( \sum{x^2y}\cdot \sum{xy^2} \right) +4xyz\cdot \sum{x^2y}\\\,\, +4xyz\cdot \sum{xy^2}\\=\left( \sum{x^4y^2}+2\sum{x^2y^3z} \right) +\left( \sum{x^2y^4}+2\sum{xy^3z^2} \right) +4x^2y^2z^2\\\,\, +2\cdot \left( \sum{x^3y^3}+\sum{x^2y^2z^2}+\sum{x^4yz} \right) +4xyz\cdot \sum{x^2y}+4xyz\cdot \sum{xy^2} 9右式=(∑x2y+∑xy2+2xyz)2=(∑x2y)2+(∑xy2)2+4x2y2z2+2⋅(∑x2y⋅∑xy2)+4xyz⋅∑x2y+4xyz⋅∑xy2=(∑x4y2+2∑x2y3z)+(∑x2y4+2∑xy3z2)+4x2y2z2+2⋅(∑x3y3+∑x2y2z2+∑x4yz)+4xyz⋅∑x2y+4xyz⋅∑xy2
经过一番整合,左式右式可以如下整理: \text{经过一番整合,左式右式可以如下整理:} 经过一番整合,左式右式可以如下整理:
∴ 左式 = 12 ∑ x 3 y 2 + 52 ∑ x 2 y 3 z + 52 ∑ x 3 y 2 z + 96 x 2 y 2 z 2 + 4 ∑ x 5 y + 4 ∑ x 5 z + 20 ∑ x 4 y z + 8 ∑ x 4 y 2 + 8 ∑ x 2 y 4 ⋯ ⋯ ⋯ ⋯ {\therefore \text{左式}=12\sum{x^3y^2}+52\sum{x^2y^3z}+52\sum{x^3y^2z}+96x^2y^2z^2+4\sum{x^5y}+4\sum{x^5z}}\\{ \,\, +20\sum{x^4yz}+8\sum{x^4y^2}+8\sum{x^2y^4}\cdots \cdots \cdots \cdots } ∴左式=12∑x3y2+52∑x2y3z+52∑x3y2z+96x2y2z2+4∑x5y+4∑x5z+20∑x4yz+8∑x4y2+8∑x2y4⋯⋯⋯⋯
右式 = 9 ∑ x 4 y 2 + 9 ∑ x 2 y 4 + 18 ∑ x 2 y 3 z + 18 ∑ x y 3 z 2 + 90 x 2 y 2 z 2 + 18 ∑ x 3 y 3 + 18 ∑ x 4 y z + 2 x y z ⋅ ( 18 ∑ x 2 y + 18 ∑ x y 2 ) ⋯ ⋯ ⋯ ⋯ { \text{右式}=9\sum{x^4y^2}+9\sum{x^2y^4}+18\sum{x^2y^3z}+18\sum{xy^3z^2}+90x^2y^2z^2+18\sum{x^3y^3}}\\{ \,\, +18\sum{x^4yz}+2xyz\cdot \left( 18\sum{x^2y}+18\sum{xy^2} \right) \cdots \cdots \cdots \cdots } 右式=9∑x4y2+9∑x2y4+18∑x2y3z+18∑xy3z2+90x2y2z2+18∑x3y3+18∑x4yz+2xyz⋅(18∑x2y+18∑xy2)⋯⋯⋯⋯
右 − 左 = − 6 ∑ x 3 y 3 − 2 x y z ⋅ ( ∑ x y 2 + ∑ x 2 y ) + 6 x 2 y 2 z 2 + 3 ∑ x 5 y + 3 ∑ x 5 z + ∑ x 5 y + ∑ x 5 z + 2 ∑ x 4 y z − ( ∑ x 4 y 2 + ∑ x 2 y 4 ) = A + B + C 右-左=-6\sum{x^3y^3}{ -2xyz\cdot \left( \sum{xy^2}+\sum{x^2y} \right) +6x^2y^2z^2}{+3\sum{x^5y}+3\sum{x^5z}}\\\,\, +\sum{x^5y}+\sum{x^5z}{+2\sum{x^4yz}}{ -\left( \sum{x^4y^2}+\sum{x^2y^4} \right) }=A+B+C 右−左=−6∑x3y3−2xyz⋅(∑xy2+∑x2y)+6x2y2z2+3∑x5y+3∑x5z+∑x5y+∑x5z+2∑x4yz−(∑x4y2+∑x2y4)=A+B+C
A = 2 x y z ⋅ [ ∑ x 3 − ∑ ( x 2 y + y 2 x ) + 3 x y z ] ≥ 0 ⋯ ⋯ ⋯ ⋯ 舒尔不等式 B = ∑ ( x 5 y + y 5 x ) − ∑ ( x 4 y 2 + x 2 y 4 ) ≥ 0 ⇔ ∑ x y ( x − y ) 2 ( x 2 + x y + y 2 ) ≥ 0 ⋯ ⋯ ⋯ ⋯ x , y , z > 0 C = 3 ⋅ [ ∑ ( x 5 y + y 5 x ) − 2 ∑ x 3 y 3 ] ≥ 0 ⋯ ⋯ ⋯ ⋯ 均值不等式 \text{A}={ 2xyz\cdot \left[ \sum{x^3}-\sum{\left( x^2y+y^2x \right)}+3xyz \right] }\ge 0\cdots \cdots \cdots \cdots \text{舒尔不等式}\\\text{B}={\sum{\left( x^5y+y^5x \right)}-\sum{\left( x^4y^2+x^2y^4 \right)}}\ge 0\\\Leftrightarrow \sum{xy\left( x-y \right) ^2\left( x^2+xy+y^2 \right) \ge 0\cdots \cdots \cdots \cdots x,y,z>0}\\\text{C}=3\cdot \left[ \sum{\left( x^5y+y^5x \right)}-2\sum{x^3y^3} \right] \ge 0\cdots \cdots \cdots \cdots \text{均值不等式} A=2xyz⋅[∑x3−∑(x2y+y2x)+3xyz]≥0⋯⋯⋯⋯舒尔不等式B=∑(x5y+y5x)−∑(x4y2+x2y4)≥0⇔∑xy(x−y)2(x2+xy+y2)≥0⋯⋯⋯⋯x,y,z>0C=3⋅[∑(x5y+y5x)−2∑x3y3]≥0⋯⋯⋯⋯均值不等式
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