Áp dụng AM-GM ta có \(\frac{1^2}{x}+\frac{1^2}{x}+\frac{1^2}{y}+\frac{1^2}{z}\ge\frac{\left(1+1+1+1\right)^2}{2x+y+z}\)

hay \(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\)

Tương tự : \(\frac{2}{y}+\frac{1}{x}+\frac{1}{z}\ge\frac{16}{2y+x+z}\) ; \(\frac{2}{z}+\frac{1}{x}+\frac{1}{y}\ge\frac{16}{2z+x+y}\)

Cộng theo vế : \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge16\left(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\right)\)

\(\Leftrightarrow\)\(16\left(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\right)\le16\)

\(\Leftrightarrow\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\le1\)

Với 2 số dương bất kì: ( 1 )

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)Vì x và y dương nên \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\forall x;y\)

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

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

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

Tương tự ta có: \(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)

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

Cộng vế với vế các bất đẳng thức kết hợp với điều kiện \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\) nên ta có đpcm

áp dụng bđt 1/a +1/b >= 4/(a+b) ta đc :....1/ (2x+y+z) <= 1/4(x+y)+1/4(x+z) ; 1/(2y+z+x)<=1/4(y+z)+1/4(x+y) ; 1/(2z+x+y)<=1/4(z+x)+1/4(z+y)

suy ra A (biểu thức đã cho ) <= 1/2(x+y) +1/2(y+z) +1/2(z+x)<= 1/8 . (1/x+1/y) +1/8. (1/y+1/z)+1/8(1/z+1/x) =1/8 . 2. (1/x+1/y+1/z)=1    (áp dụng lại bđt trên)...,suyra đpcm.

dấu ''='' xảy ra <=> x=y=z

câu hỏi tương tự nhé bạn

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

Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\hept{\begin{cases}\frac{1}{x+y+x+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{z+y+x+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\\\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\end{cases}}\)

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

\(\hept{\begin{cases}\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\\\frac{1}{x+z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{z}\right)\\\frac{1}{z+y}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{y}\right)\end{cases}}\Rightarrow\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{2z+x+y}+\frac{1}{2y+z+x}\le\frac{1}{2}\cdot\frac{1}{2}\cdot4=1\)

\("="\Leftrightarrow x=y=z=0,75\)

bùi huyền ơi làm sao để k cho bạn được

làm sao để k cho bạn vậy

Theo Cauche có: 

\(\left(x+x+y+z\right)\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\sqrt[4]{x^2yz}.4\sqrt[4]{\frac{1}{x^2.y.z}}=16\)

=> \(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\). Tương tự có: 

\(\frac{2}{y}+\frac{1}{x}+\frac{1}{z}\ge\frac{16}{x+2y+z}\) và \(\frac{2}{z}+\frac{1}{y}+\frac{1}{x}\ge\frac{16}{x+y+2z}\)

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

\(16.\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le4.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=4.4=16\)

Chia cả 2 vế cho 16 => ĐPCM

Áp dụng công thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x,y>0\right)\)

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

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

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

Tương tự \(\hept{\begin{cases}\frac{1}{x+2y+z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{2y}+\frac{1}{4z}\right)\left(2\right)\\\frac{1}{x+y+2z}\le\frac{1}{4}\left(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{2z}\right)\left(3\right)\end{cases}}\)

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

=> \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)

Bài này áp dụng BĐT này nhé , với x,y > 0 ta có :

\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ( Cách chứng minh thì chuyển vế quy đồng nhé )

Áp dụng vào bài toán ta có :

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

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

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

Tương tự ta có :

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

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

Do đó : \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=\frac{1}{4}\left(x+y+z\right)=1\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{3}{4}\) (đpcm)

Ta có: \(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

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

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

Cộng vế theo vế có: \(VT\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=1\)

cách 1:

với a,b>0 ta có: 4ab < (a+b)² \(\Leftrightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)

dấu "=" xảy ra khi a=b

áp dụng kết quả của trên ta có:

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

tương tự \(\hept{\begin{cases}\frac{1}{x+2y+z}\le\frac{1}{4}\left[\frac{1}{2y}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{z}\right)\right]=\frac{1}{8}\left(\frac{1}{y}+\frac{1}{2x}+\frac{1}{2z}\right)\left(2\right)\\\frac{1}{x+y+2z}\le\frac{1}{4}\left[\frac{1}{2z}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\right]=\frac{1}{8}\left(\frac{1}{z}+\frac{1}{2y}+\frac{2}{2x}\right)\left(3\right)\end{cases}}\)

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

thấy trong các bđt (1)(2)(3) thì dấu "=" xảy ra khi x=y=z=\(\frac{3}{4}\)

cách 2:

áp dụng bđt 1\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)và bđt Cosi cho các số dương ta có:

\(2x+y+z=\left(x+y\right)+\left(x+z\right)\ge2\left(\sqrt{xy}+\sqrt{xyz}\right)\)

do đó: \(\frac{1}{2x+y+z}\le\frac{1}{2}\left(\frac{1}{\sqrt{xy}+\sqrt{xz}}\right)\le\frac{1}{8}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{xz}}\right)\)

tương tự: \(\hept{\begin{cases}\frac{1}{2x+y+z}\le\frac{1}{8}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}\right)\\\frac{1}{x+y+2z}\le\frac{1}{8}\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yz}}\right)\end{cases}}\)

cộng theo từng vế 3 bđt trên ta được:

\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\left(3\right)\)

từ (3), (4) => đpcm

cách 3:

mặt khác từ bđt Cosi cho 4 số dương hoặc bđt Bunhiacopsky

\(\left(x+x+y+z\right)\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\sqrt[4]{x^2\cdot yz}\ge4\sqrt[4]{\frac{1}{x^2yz}}=16\)

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

tương tự \(\hept{\begin{cases}\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\\\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\end{cases}}\)

cộng 3 vế của bđt trên ta được đpcm

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

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

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

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

\("="\Leftrightarrow x=y=z=\frac{3}{4}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:

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

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

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

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

\(''=''\Leftrightarrow x=y=z=\frac{3}{4}\)

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

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

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

Do đó : \(2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)

\(\Rightarrow\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\le2\)

+ \(\frac{1}{x+y}+\frac{1}{y+z}\ge\frac{4}{x+2y+z}\).Dấu "=" xảy ra\(\Leftrightarrow x=z\)

\(\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{4}{x+y+2z}\). Dấu "=" xảy ra \(\Leftrightarrow x=y\)

\(\frac{1}{x+y}+\frac{1}{z+x}\ge\frac{4}{2x+y+z}\). Dấu "=" xảy ra \(\Leftrightarrow y=z\)

Do đó : \(2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge4\left(\frac{1}{x+2y+z}+\frac{1}{x+y+2z}+\frac{1}{2x+y+z}\right)\)

=> đpcm

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{3}{4}\)