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23 tháng 3 2020

Đặt \(a=2x+y+z;b=2y+z+x;c=2z+x+y\)

\( \implies\) \(a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)\) 

\( \implies\) \(a+b+c=4x+4y+4z\)

\( \implies\) \(x+y+z=\frac{a+b+c}{4}\) 

+)Ta có : \(a=2x+y+z\)

\(\iff\) \(a=x+\left(x+y+z\right)\)

\(\iff\) \(a-\left(x+y+z\right)=x\)

\(\iff\) \(a-\frac{a+b+c}{4}=x\)

\(\iff\) \(x=\frac{3a-b-c}{4}\)

+)Ta có :\(b=2y+z+x\)

\(\iff\) \(b=y+\left(y+z+x\right)\)

\(\iff\)\(b-\left(y+z+x\right)=y\)

\(\iff\) \(b-\frac{a+b+c}{4}=y\)

\(\iff\)\(y=\frac{3b-c-a}{4}\)

+)Ta có :\(c=2z+x+y\)

\(\iff\) \(c=z+\left(z+x+y\right)\)

\(\iff\) \(c-\left(z+x+y\right)=z\)

\(\iff\) \(c-\frac{a+b+c}{4}=z\)

\(\iff\)\(z=\frac{3c-a-b}{4}\)

​​\( \implies\)​ \(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\) 

 \(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)

 \(=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)\)

 \(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)

 \(=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\)

Áp dụng bất đẳng thức ( BĐT Cosi ) : \(m+n\)\( \geq\)\(2\sqrt{mn}\) \(\left(m;n>0\right)\)ta được : 

\(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{a}.\frac{a}{b}}\) = 2 \( \implies\) \(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 

\(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 \(\sqrt{\frac{c}{a}.\frac{a}{c}}\) = 2 \( \implies\) \(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 

\(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{c}.\frac{c}{b}}\) = 2 \( \implies\) \(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 

\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 2 + 2 + 2 

\( \implies\) ​​\(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)​ \( \geq\) 6 

\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)

\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)

\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)

\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)

\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\) 

23 tháng 3 2020

Dấu " = " xảy ra khi a = b = c hay x = y = z 

9 tháng 3 2019

Áp dụng bđt Cauchy-Schwarz:

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

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

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

\("="\Leftrightarrow x=y=z\)

20 tháng 4 2015

đặt a = 2x + y + z; b = 2y + z + x; c = 2z + x + y (a; b ; c > 0)

=> a + b + c = 4.(x+ y + z) => x + y + z = (a+ b+ c) / 4

=> x = a - (x+ y + z) = a - (a+ b + c) / 4 

y = b - (x + y + z) = b - (a+b+c) / 4

z = c - (x+y + z) = c - (a+b+c)/ 4 

Khi đó :  \(VT=1-\frac{a+b+c}{4a}+1-\frac{a+b+c}{4b}+1-\frac{a+b+c}{4c}\)

\(VT=3-\left(\frac{a+b+c}{4a}+\frac{a+b+c}{4b}+\frac{a+b+c}{4c}\right)=3-\frac{1}{4}.\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(VT=3-\frac{1}{4}.\left(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\right)=3-\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\)

Với a, b > 0 ta có: a/b + b/ a > = 2

=> \(\frac{1}{4}.\left(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right)\ge\frac{1}{4}.\left(3+2+2+2\right)=\frac{9}{4}\)

=> \(VT\le3-\frac{9}{4}=\frac{3}{4}\)

Dấu = xảy ra khi a= b = c => x = y = z 

19 tháng 7 2020

Theo Cauchy Schwarz:

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

Tương tự:

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

Cộng lại:

\(D\le\frac{3}{4}\left(đpcm\right)\)

20 tháng 7 2017

Cậu vào đây nha ! 

Câu hỏi của doanthihuong - Toán lớp 7 - Học toán với OnlineMath

25 tháng 7 2018

Đặt \(\hept{\begin{cases}2x+y+z=4a\\2y+x+z=4b\\2z+x+y=4c\end{cases}\Rightarrow}\hept{\begin{cases}x=3a-b-c\\y=3b-c-a\\z=3c-a-b\end{cases}}\)thay vào biểu thức đó

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

\(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)

\(=\frac{3}{4}-\frac{b-c}{4a}+\frac{3}{4}-\frac{c-a}{4b}+\frac{3}{4}-\frac{a-b}{4c}\)

\(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)

Áp dụng BĐT sau: \(\frac{a}{b}+\frac{b}{a}\ge2\Rightarrow\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\ge6\)

\(\Leftrightarrow\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\ge\frac{6}{4}\)

\(\Leftrightarrow\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\le\frac{3}{4}\)

Từ đó ta có: \(\frac{x}{2x+y+z}+\frac{y}{2y+x+z}+\frac{z}{2z+x+y}\le\frac{3}{4}\)(đpcm).

Dấu "=" xảy ra <=> x=y=z.