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

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

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

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

Áp dụng bất đẳng thức : \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)( với x , y > 0 )
Ta có : \(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{2x}+\frac{1}{y+z}\right);\frac{1}{y+z}\le\frac{1}{4y}+\frac{1}{4z}\)

Suy ra : 

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

Tường tự ta có : 

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

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

Từ (1) , (2) và (3) 

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

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

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

Chúc bạn học tốt !!!

địt mẹ laaaaaa

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 <=> \(x=y=z=\frac{3}{4}\)

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

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

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

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

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

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

3) Đặt b+c=x;c+a=y;a+b=z.

=>a=(y+z-x)/2 ; b=(x+z-y)/2 ; c=(x+y-z)/2

BĐT cần CM <=> \(\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)

VT=\(\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}-1+\frac{x}{y}+\frac{z}{y}-1+\frac{x}{z}+\frac{y}{z}-1\right)\)

\(=\frac{1}{2}\left[\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)-3\right]\)

\(\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)(Cauchy)

Dấu''='' tự giải ra nhá

Bài 4 

dễ chứng minh \(\left(a+b\right)^2\ge4ab;\left(b+c\right)^2\ge4bc;\left(a+c\right)^2\ge4ac\)

\(\Rightarrow\left(a+b\right)^2\left(b+c\right)^2\left(a+c\right)^2\ge64a^2b^2c^2\)

rồi khai căn ra \(\Rightarrow\)dpcm. 

đấu " = " xảy ra \(\Leftrightarrow\)\(a=b=c\)

bài 1 \(\left(\frac{x}{y}\right)^2+\left(\frac{y}{z}\right)^2\ge2\times\frac{x}{y}\times\frac{y}{z}=2\frac{x}{z}\)

làm tương tự rồi cộng các vế các bất đẳng thức lại với nhau ta có dpcm ( cộng xong bạn đặt 2 ra ngoài ý, mk ngại viết nhiều hhehe) 

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

Giải:

Ta có: x, y, z >0

Áp dụng BĐT Cô si ta có:

\(\left(x+y\right)\ge2\sqrt{xy}\) và \(\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{\frac{1}{xy}}\)

=> \(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{xy}.2\sqrt{\frac{1}{xy}}=4\)

<=> \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\frac{1}{x+y}\le4\left(\frac{1}{x}+\frac{1}{y}\right)\)               (*)

Áp dụng (*) ta có: 

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

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

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

Cộng 2 vế của (1), (2), (3) ta có

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

cảm ơn bạn nhiều