cho x,y,z là các số dương thỏa mãn :\(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=6\)
chứng minh \(\dfrac{1}{3x+3y+2z}+\dfrac{1}{3x+2y+3z}+\dfrac{1}{2x+2y+3z}\le\dfrac{3}{2}\)
K
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18 tháng 12 2021
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
17 tháng 7 2021
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
10 tháng 11 2017
Ta có :
\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(2x+y+z\right)+\left(2y+x+z\right)}\)(1)
Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\left(1\right)\le\dfrac{1}{4}\left(\dfrac{1}{x+y+x+z}+\dfrac{1}{y+x+y+z}\right)\le\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\right)\)
\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
tương tự với hai ông còn lại sau đó cộng lại ta được:
\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 3 2021
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
KK
13 tháng 7 2017
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)
Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)
16 tháng 7 2017
Trai Vô Đối câu này đề thi vô lớp 10 tỉnh Thanh Hóa ( tất cả thí sinh nek .... lúc nào rảnh mik đăng lên thử xem sao )
AH
Akai Haruma
Giáo viên
17 tháng 9 2017
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\geq \frac{16}{3x+3y+2z}\)
\(\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\geq \frac{16}{3x+2y+3z}\)
\(\frac{1}{z+y}+\frac{1}{z+y}+\frac{1}{x+z}+\frac{1}{x+y}\geq \frac{16}{2x+3y+3z}\)
Cộng theo vế:
\(\Rightarrow 4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\geq 16\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)
\(\Rightarrow \frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\leq \frac{4.6}{16}=\frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
23 tháng 4 2017
Haha không giỡn nữa :v 
Áp dụng BĐT Cauchy-Schwarz ta có:
\(L.H.S=Σ\dfrac{1}{2x+y+z}=7Σ\dfrac{1}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)
\(=\dfrac{1}{7}Σ\dfrac{\left(2+1+4\right)^2}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)
\(\le\dfrac{1}{7}Σ\left(\dfrac{2^2}{2\left(x+3y\right)}+\dfrac{1^2}{y+3z}+\dfrac{4^2}{4\left(z+3x\right)}\right)\)
\(=\dfrac{1}{7}Σ\left(\dfrac{2}{x+3y}+\dfrac{1}{y+3z}+\dfrac{4}{z+3x}\right)\)
\(=\dfrac{1}{7}Σ\dfrac{7}{x+3y}=Σ\dfrac{1}{x+3y}=R.H.S\)
KK
23 tháng 4 2017
Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\le\dfrac{4}{x+y}\) \(\forall x,y>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+3y}+\dfrac{1}{y+2z+x}\le\dfrac{4}{2x+4y+2z}=\dfrac{2}{x+2y+z}\\\dfrac{1}{y+3z}+\dfrac{1}{z+2x+y}\le\dfrac{4}{2x+2y+4z}=\dfrac{2}{x+y+2z}\\\dfrac{1}{z+3x}+\dfrac{1}{x+2y+z}\le\dfrac{4}{4x+2y+2z}=\dfrac{2}{2x+y+z}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}+\dfrac{1}{y+2z+x}+\dfrac{1}{z+2x+y}+\dfrac{1}{x+2y+z}\le\dfrac{2}{x+2y+z}+\dfrac{2}{x+y+2z}+\dfrac{2}{2x+y+z}\)
\(\Rightarrow VT\le\left(\dfrac{2}{x+2y+z}-\dfrac{1}{x+2y+z}\right)+\left(\dfrac{2}{x+y+2z}-\dfrac{1}{y+x+2z}\right)+\left(\dfrac{2}{2x+y+z}-\dfrac{1}{z+2x+y}\right)\)
\(\Rightarrow VT\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\)
\(\Leftrightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\) ( đpcm )
Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)
CMTT và cộng theo vế:
\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)