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4 tháng 10 2016

Áp dụng Bđt \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

Ta có:

\(\frac{1}{2x+3y+3z}=\frac{1}{\left(x+2y+z\right)+\left(x+y+2z\right)}\)\(\le\frac{1}{4}\left(\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)

\(=\frac{1}{4}\cdot\left(\frac{1}{\left(x+y\right)+\left(y+z\right)}+\frac{1}{x+z}+\frac{1}{z+y}\right)\)

\(\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\right]+\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

\(=\frac{1}{16}\left(6+\frac{1}{y+z}\right)\).Tương tự với 2 cái còn lại r` cộng lại ta đc:

\(P\le\frac{1}{16}\left[6+6+6+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right]=\frac{3}{2}\)

NV
15 tháng 4 2019

Đặt \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\x+z=c\end{matrix}\right.\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

\(P=\frac{1}{a+a+b+c}+\frac{1}{a+b+b+c}+\frac{1}{a+b+c+c}\)

\(\Rightarrow P\le\frac{1}{16}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)+\frac{1}{16}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)

\(\Rightarrow P\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)

\(\Rightarrow P_{max}=\frac{3}{2}\) khi \(a=b=c=\frac{1}{2}\Rightarrow x=y=z=\frac{1}{4}\)

NV
4 tháng 5 2020

\(\frac{1}{2x+3y+3z}=\frac{1}{\left(x+y\right)+\left(x+z\right)+\left(y+z\right)+\left(y+z\right)}\le\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{2}{y+z}\right)\)

Tương tự:

\(\frac{1}{3x+2y+3z}\le\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{2}{x+z}\right)\) ; \(\frac{1}{3x+3y+2z}\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\)

Cộng vế với vế:

\(P\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{2017}{4}\)

Dấu "=" xảy ra khi \(x=y=z=\frac{3}{4034}\)

1 tháng 2 2018

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\)ta có :

\(\frac{16}{3x+3y+2z}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)

\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\)

\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\)

Cộng theo vế 3 đẳng thức trên ta được :

\(16.\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)

\(\le4.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=4.6=24\)

\(\Rightarrow\)\(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\)

1 tháng 2 2018

Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath

13 tháng 10 2019

Câu 1:

\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)

\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)

Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)

\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)

Dấu = xảy ra khi x=y=1/2

13 tháng 10 2019

Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)

CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu = xảy ra khi x=y=z=1

21 tháng 10 2019

Liên tục áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) và ta có:

\(\frac{1}{3x+3y+2x}=\frac{1}{2\left(x+y\right)+\left(x+y+2z\right)}\le\frac{1}{4}\left(\frac{1}{2\left(x+y\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\right)\le\frac{1}{8\left(x+y\right)}+\frac{1}{16}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

Chứng minh tương tự tạ có:

\(\frac{1}{3x+2y+3z}\le\frac{1}{8\left(z+x\right)}+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

\(\frac{1}{2x+3y+3z}\le\frac{1}{8\left(y+z\right)}+\frac{1}{16}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)\)

Suy ra \(VT\le\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)+\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{4}\)

21 tháng 10 2019

mơn ạ

9 tháng 1 2019

Ta có bđt \(\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)\)

\(\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)\)

Áp dụng nhiều lần bđt trên ta được

\(\(\frac{1}{3x+3y+2z}=\frac{1}{\left(2x+y+z\right)+\left(x+2y+z\right)}\le\frac{1}{4}\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}\right)\)\)

\(\(\le\frac{1}{4}\left(\frac{1}{\left(x+y\right)+\left(x+z\right)}+\frac{1}{\left(x+y\right)+\left(y+z\right)}\right)\)\)

\(\(\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\right]\)\)

\(\(\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)\)

C/m tương tự cho các bđt còn lại

\(\(\frac{1}{3x+2y+3z}\le\frac{1}{16}\left(\frac{2}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\right)\)\)

\(\(\frac{1}{2x+3y+3z}\le\frac{1}{16}\left(\frac{2}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\right)\)\)

Cộng vế theo vế được

\(\(P\le\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)=\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{1}{4}.6=\frac{3}{2}\)\)

Dấu "=" xảy ra

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{1}{2x}+\frac{1}{2x}+\frac{1}{2x=6}\end{cases}}\)\)

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\\frac{3}{2x}=6\end{cases}}\)\)

\(\(\Leftrightarrow\hept{\begin{cases}x=y=z\\x=\frac{1}{4}\end{cases}}\)\)

\(\(\Leftrightarrow x=y=z=\frac{1}{4}\)\)

Vậy ..........

10 tháng 1 2019

cách khác :)) 

\(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\)\(\Leftrightarrow\)\(x+y+z\le3\)

\(P=\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\)

\(P=\frac{1}{3\left(x+y+z\right)-z}+\frac{1}{3\left(x+y+z\right)-y}+\frac{1}{3\left(x+y+z\right)-x}\)

\(\ge\frac{9}{9\left(x+y+z\right)-\left(x+y+z\right)}=\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.3}=\frac{3}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{4}\)

26 tháng 10 2016

Ta có:

\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\ge\frac{9}{2\left(x+y+z\right)}\)\(\Rightarrow x+y+z\ge\frac{3}{4}\)

Lại có: \(\frac{1}{2x+3y+3z}=\frac{\left(\frac{3}{4}+\frac{1}{4}\right)^2}{2\left(x+y+z\right)+y+z}\le\frac{9}{32\left(x+y+z\right)}+\frac{1}{16\left(y+z\right)}\)

Do đó:

\(\frac{1}{2x+3y+3z}+\frac{1}{2y+3x+3z}+\frac{1}{2z+3x+3y}\)

\(\le\frac{9}{32\left(x+y+z\right)}\cdot3+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

\(\le\frac{9}{32\cdot\frac{3}{4}}+\frac{1}{16}\cdot6=\frac{3}{2}\)(Đpcm)

1 tháng 2 2018

Tại sao \(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=6\ge\frac{9}{2\left(x+y+z\right)}\)

13 tháng 10 2019

\(\text{Áp dụng BĐT:}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)

\(\frac{1}{3x+3y+2z}=\frac{1}{\left(x+y\right)+\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)

\(\text{tương tự với các BĐT còn lại }\)

\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+3z+2y}+\frac{1}{3y+3z+2x}\le\frac{1}{16}.\left(\frac{4}{x+z}+\frac{4}{x+y}+\frac{4}{y+z}\right)=\frac{1}{16}.24=\frac{3}{2}\left(đpcm\right)\)

NV
27 tháng 4 2020

Đặt \(\left(y+z;x+z;x+y\right)=\left(a;b;c\right)\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{b+c-a}{2}\\y=\frac{a+c-b}{2}\\z=\frac{a+b-c}{2}\end{matrix}\right.\)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2017}\)

\(P=\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\)

\(P\le\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{2}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\right)\)

\(P\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{2017}{4}\)

Dấu "=" xảy ra khi \(a=b=c=6051\) hay \(x=y=z=\frac{6051}{2}\)