\(\Leftrightarrow\left(x^2y-2xyz+z^2y\right)+\left(x^2z-y^2x-z^2x+y^2z\right)=0\)

\(\Leftrightarrow y\left(x-z\right)^2+xz\left(x-z\right)-y^2\left(x-z\right)=0\)

\(\Leftrightarrow\left(x-z\right)\left(xy-yz+zx-y^2\right)=0\)

\(\Leftrightarrow\left(x-z\right)\left(x\left(y+z\right)-y\left(y+z\right)\right)=0\)

\(\Leftrightarrow\left(x-z\right)\left(x-y\right)\left(y+z\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\) hay có 2 số bằng hoặc đối nhau

(x² y - y² x) + (x² z - xyz) + (z² y - z² x) + (y² z - xyz) = (x-y)(xy+zx-z² -yz)=(x-y)(x-z)(y+z)=0

Giải giùm rồi đấy bạn

Giải giùm mik nha!

bạn nhiều câu hỏi quá

\(x^2y-xy^2+x^2z-xz^2+y^2z+yz^2=2xyz\)

\(\Leftrightarrow\left(x^2y-xy^2\right)+\left(x^2z-xyz\right)-\left(xz^2-yz^2\right)-\left(xyz-y^2z\right)=0\)

\(\Leftrightarrow xy\left(x-y\right)+xz\left(x-y\right)-z^2\left(x-y\right)-yz\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(xy+xz-z^2-yz\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[x\left(y+z\right)-z\left(y+z\right)\right]=0\)

\(\Leftrightarrow\left(x-y\right)\left(x-z\right)\left(y+z\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=z\\y=-z\end{matrix}\right.\)\(\left(đpcm\right)\)

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 

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

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

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

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

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

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

BĐT tương đương:

\(\frac{1}{z\left(1+\frac{1}{x}\right)}+\frac{1}{x\left(1+\frac{1}{y}\right)}+\frac{1}{y\left(1+\frac{1}{z}\right)}\ge2\)

Từ giả thiết:

\(xy+yz+zx+2xyz=1\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2=\frac{1}{xyz}\)

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a+b+c+2=abc\)

\(\Rightarrow a+b+c+2\le\frac{1}{27}\left(a+b+c\right)^3\)

\(\Leftrightarrow\left(a+b+c\right)^3-27\left(a+b+c\right)-54\ge0\)

\(\Leftrightarrow\left(a+b+c-6\right)\left(a+b+c+3\right)^2\ge0\)

\(\Leftrightarrow a+b+c\ge6\)

BĐT trở thành: \(\frac{c}{1+a}+\frac{a}{1+b}+\frac{b}{1+c}\ge2\)

Thật vậy, ta có:

\(VT=\frac{a^2}{a+ab}+\frac{b^2}{b+bc}+\frac{c^2}{c+ca}\ge\frac{\left(a+b+c\right)^2}{a+b+c+ab+bc+ca}\ge\frac{3\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(a+b+c\right)^2}\)

\(VT\ge\frac{3\left(a+b+c\right)}{3+a+b+c}=\frac{2\left(a+b+c\right)+a+b+c}{a+b+c+3}\ge\frac{2\left(a+b+c\right)+6}{a+b+c+3}=2\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=2\) hay \(x=y=z=\frac{1}{2}\)

Đặt \(\hept{\begin{cases}\frac{1}{x^2}=a\\\frac{1}{y^2}=b\\\frac{1}{z^2}=c\end{cases}}\Rightarrow abc=1\) và ta cần chứng minh 

\(\frac{1}{2a+b+3}+\frac{1}{2b+c+3}+\frac{1}{2c+a+3}\le\frac{1}{2}\left(1\right)\)

Áp dụng BĐT AM-GM ta có: 

\(2a+b+3=\left(a+b\right)+\left(a+1\right)+2\ge2\left(\sqrt{ab}+\sqrt{a}+2\right)\)

\(\Rightarrow\frac{1}{2a+b+3}\le\frac{1}{2\left(\sqrt{ab}+\sqrt{a}+1\right)}=\frac{1}{2}\cdot\frac{1}{\sqrt{ab}+\sqrt{a}+1}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{1}{2b+c+3}\le\frac{1}{2}\cdot\frac{1}{\sqrt{bc}+\sqrt{b}+1};\frac{1}{2c+a+3}\le\frac{1}{2}\cdot\frac{1}{\sqrt{ac}+\sqrt{c}+1}\)

Cộng theo vế 3 BĐT trên ta có: 

\(VT_{\left(1\right)}\le\frac{1}{2}\left(\frac{1}{\sqrt{ab}+\sqrt{a}+1}+\frac{1}{\sqrt{b}+\sqrt{bc}+1}+\frac{1}{\sqrt{c}+\sqrt{ac}+1}\right)\le\frac{1}{2}=VP_{\left(2\right)}\left(abc=1\right)\)

t nghĩ ôg có chút nhầm lẫn , phải là sigma (1/2b+a+3) </ 1/2 

\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)

\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\) 

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)

bạn ơi hình như có chút sai đề