Ta có: \(x^2+y^2\ge2xy\)

\(\sqrt{x^2+y^2}\ge\sqrt{\frac{\left(x+y\right)^2}{2}}=\frac{\sqrt{2}}{2}\left(x+y\right)\)

Do các vế của BĐT đều dương, nhân vế với vế:

\(\left(x^2+y^2\right)\sqrt{x^2+y^2}\ge\sqrt{2}xy\left(x+y\right)\) (đpcm)

Dấu "=" xảy ra khi \(x=y\)

Please !!!!!

Tương tự, ta được:

\(\left(2-y\right)\left(2-z\right)>=\dfrac{\left(x+1\right)^2}{4}\)

và \(\left(2-z\right)\left(2-x\right)>=\left(\dfrac{y+1}{2}\right)^2\)

=>8(2-x)(2-y)(2-z)>=(x+1)(y+1)(z+1)

(x+yz)(y+zx)<=(x+y+yz+xz)^2/4=(x+y)^2*(z+1)^2/4<=(x^2+y^2)(z+1)^2/4

Tương tự, ta cũng co:

\(\left(y+xz\right)\left(z+y\right)< =\dfrac{\left(y^2+z^2\right)\left(x+1\right)^2}{2}\)

và \(\left(z+xy\right)\left(x+yz\right)< =\dfrac{\left(z^2+x^2\right)\left(y+1\right)^2}{2}\)

Do đó, ta được:

\(\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)< =\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

=>ĐPCM

a.

\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)

b.

\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)

TH1:

\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)

Xét pt:

\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)

\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)

TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)

\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)

\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)

TH2: \(y=6-x\) thế vào...

\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)

\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)

Dễ thấy:

     \(VT\ge\left(x+y\right)^2+1-\dfrac{\left(x+y\right)^2}{4}=\dfrac{3\left(x+y\right)^2}{4}+1\)

Áp dụng Cô-si:

     \(\dfrac{3\left(x+y\right)^2}{4}+1\ge2\sqrt{\dfrac{3\left(x+y\right)^2}{4}.1}=\sqrt{3}\left|x+y\right|\ge\sqrt{3}\left(x+y\right)\)

Do đó:

     \(\left(x+y\right)^2+1-xy\ge\sqrt{3}\left(x+y\right),\forall x,y\in R\)

\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

Dâu "=" xảy ra khi \(x=y=z\)