may ban cho mk hoi bai nay:cho x>1,y>1,z>1 c/m:2xyz+1>xy+yz+xz
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
9 tháng 4 2017
\(P=\dfrac{xy}{1+x+y}+\dfrac{yz}{1+y+z}+\dfrac{xz}{1+z+x}\)
\(P+3=\dfrac{xy}{1+x+y}+1+\dfrac{yz}{1+y+z}+1+\dfrac{xz}{1+z+x}+1\)
\(P+3=\dfrac{\left(x+1\right)\left(y+1\right)}{1+x+y}+\dfrac{\left(y+1\right)\left(z+1\right)}{1+y+z}+\dfrac{\left(x+1\right)\left(z+1\right)}{1+z+x}\)
\(P+3=\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(1+x+y\right)\left(z+1\right)}+\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(x+1\right)\left(1+y+z\right)}+\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(y+1\right)\left(1+z+x\right)}\)
\(P+3=\left(x+1\right)\left(y+1\right)\left(z+1\right)\left[\dfrac{1}{\left(1+x+y\right)\left(z+1\right)}+\dfrac{1}{\left(x+1\right)\left(1+y+z\right)}+\dfrac{1}{\left(y+1\right)\left(1+z+x\right)}\right]\)
\(\ge\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\left(1+x+y\right)\left(z+1\right)+\left(x+1\right)\left(1+y+z\right)+\left(y+1\right)\left(1+z+x\right)}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2xy+2yz+2xz+3x+3y+3z+3}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2xy+2yz+2xz+3\cdot2xyz}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2\left(xy+yz+xz+3xyz\right)}\)
Lại có:
\(\left(x+1\right)\left(y+1\right)\left(z+1\right)=xyz+xy+yz+xz+x+y+z+1\)
\(=xyz+xy+yz+xz+2xyz=xy+yz+xz+3xyz\)
\(\Rightarrow P+3\ge\left(xy+yz+xz+3xyz\right)\cdot\dfrac{9}{2\left(xy+yz+xz+3xyz\right)}\)
\(\Rightarrow P+3\ge\dfrac{9}{2}\Rightarrow P\ge\dfrac{9}{2}-3=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1+\sqrt{3}}{2}\)
AJ
1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2020
Bài này x;y;z phải dương chứ nhỉ? Có dấu "=" ở số 0 thế kia thì bối rối quá
Dự đoán dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
Theo nguyên lý Dirichlet, trong 3 số x;y;z luôn tồn tại 2 số nằm cùng phía so với \(\frac{1}{2}\) ; giả sử đó là x và y
\(\Rightarrow\left(x-\frac{1}{2}\right)\left(y-\frac{1}{2}\right)\ge0\Leftrightarrow\frac{1}{2}\left(x+y\right)-xy\le\frac{1}{4}\)
\(\Leftrightarrow x+y-2xy\le\frac{1}{2}\)
Mặt khác:
\(1=2xyz+x^2+y^2+z^2\ge2xyz+2xy+z^2=2xy\left(1+z\right)+z^2\)
\(\Rightarrow1-z^2\ge2xy\left(1+z\right)\Leftrightarrow\left(1-z\right)\left(1+z\right)\ge2xy\left(1+z\right)\)
\(\Leftrightarrow1-z\ge2xy\Rightarrow xy\le\frac{1-z}{2}\)
\(\Rightarrow P=xy+z\left(x+y-2xy\right)\le\frac{1-z}{2}+\frac{z}{2}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
AP
0
NL
1
HT
10 tháng 8 2016
\(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz.\)
\(=x^2y+xy^2+y^2z+yz^2+xz\left(x+z\right)+2xyz\)
\(=\left(x^2y+xyz\right)+\left(xy^2+y^2z\right)+\text{(}yz^2+xyz\text{)}+xz\left(x+z\right)\)
\(=xy\left(x+z\right)+y^2\left(x+z\right)+yz\left(x+z\right)+xz\left(x+z\right)\)
\(=\left(x+z\right)\left(xy+y^2+yz+xz\right)\)
\(=\left(x+z\right)\text{[}y\left(x+y\right)+z\left(x+y\right)\text{]}\)
\(=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)
NP
1
24 tháng 2 2020
Ta có:
P=\(\left(X^2+y^2+z^2+2xyz\right)-\left(X^2+y^2+z^2+4xyz-xy-yz-xz\right)\) xz)
= 1-\(\left(x^2+y^2+z^2+4xyz-xy-yz-xz\right)\)
=> P \(\le\)1
Vậy MaxP=1
HF
3 tháng 10 2019
\(\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\)
\(\le\frac{1}{2\sqrt{x^2yz}}+\frac{1}{2\sqrt{y^2xz}}+\frac{1}{2\sqrt{z^2xy}}=\frac{\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}}{2\sqrt{xyz}}\)
\(=\frac{\sqrt{yz}+\sqrt{xz}+\sqrt{xy}}{2xyz}\le\frac{\frac{x+y+x+z+x+y}{2}}{2xyz}=\frac{x+y+z}{2xyz}\)
Dấu '=' xảy ra <=> x=y=z
NN
14 tháng 10 2020
1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)
\(=4x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(4x+1\right)\)
2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)
\(=4x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(4x-3\right)\)
3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)
\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)
4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=xy\left(x+y\right)-y^2z-yz^2+x^2z-xz^2\)
\(=xy\left(x+y\right)+\left(x^2z-y^2z\right)-\left(yz^2+xz^2\right)\)
\(=xy\left(x+y\right)+z\left(x^2-y^2\right)-z^2.\left(x+y\right)\)
\(=xy\left(x+y\right)+z\left(x-y\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=xy\left(x+y\right)+\left(zx-zy\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)
14 tháng 10 2020
1) 4x2 - 7x - 2 = 4x2 - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )
2) 4x2 + 5x - 6 = 4x2 - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )
3) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
4) xy( x + y ) - yz( y + z ) + xz( x - z )
= x2y + xy2 - y2z - yz2 + xz( x - z )
= ( x2y - yz2 ) + ( xy2 - y2z ) + xz( x - z )
= y( x2 - z2 ) + y2( x - z ) + xz( x - z )
= y( x - z )( x + z ) + y2( x - z ) + xz( x - z )
= ( x - z )[ y( x + z ) + y2 + xz ]
= ( x - z )( xy + yz + y2 + xz )
= ( x - z )[ ( xy + y2 ) + ( xz + yz ) ]
= ( x - z )[ y( x + y ) + z( x + y ) ]
= ( x - z )( x + y )( y + z )
5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 6 2020
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)>0\Rightarrow a+b+c=2\)
\(\Rightarrow P=\frac{a^3}{\left(2-a\right)^2}+\frac{b^3}{\left(2-b\right)^2}+\frac{c^3}{\left(2-c\right)^2}\)
Ta có đánh giá: \(\frac{a^3}{\left(2-a\right)^2}\ge\frac{2a-1}{2}\) ; \(\forall a\in\left(0;2\right)\)
Thật vậy, BĐT tương đương:
\(2a^3\ge\left(2a-1\right)\left(a^2-4a+4\right)\)
\(\Leftrightarrow9a^2-12a+4\ge0\Leftrightarrow\left(3a-2\right)^2\ge0\) (luôn đúng)
Tương tự: \(\frac{b^3}{\left(2-b\right)^2}\ge\frac{2b-1}{2}\) ; \(\frac{c^3}{\left(2-c\right)^2}\ge\frac{2c-1}{2}\)
Cộng vế với vế: \(P\ge\frac{2\left(a+b+c\right)-3}{2}=\frac{1}{2}\)
\(P_{min}=\frac{1}{2}\) khi \(a=b=c=\frac{2}{3}\) hay \(x=y=z=\frac{3}{2}\)
KK
0