\(x^2+y^2-xy\ge x+y-1\)

\(\Leftrightarrow2x^2+2y^2-2xy\ge2x+2y-2\)

\(\Leftrightarrow2x^2+2y^2-2xy-2x-2y+2\ge0\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(x-y\right)^2\ge0\)

Bat ddang thuc cuoiđung,cac phep biendddooii tren la tuong dduong nen BĐT cuoi ddung =>đpcm

xay ra--ddang--thuc khi x=y=1

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

Áp dụng BĐT cô si ,ta có:

\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x\cdot y}{y\cdot x}}=2\)

Vậy ta được đpcm

ta có:

\(a+\frac{1}{a}-2=\left(\sqrt{a}\right)^2+\left(\frac{1}{\sqrt{a}}\right)^2-2\sqrt{a\cdot\frac{1}{a}}=\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)^2\ge0\Rightarrow a+\frac{1}{a}\ge2\)

Vì a và 1/a cùng dấu nên 2 căn (a*1/a) lớn hơn 0 nha 

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

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

\(=27-6\left(x+y+z\right)+x^2+y^2+z^2\)

\(=9+x^2+y^2+z^2\)

Dễ dàng CM được \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=3\)

=>\(2\left(x^2+y^2+z^2+xy+yz+zx\right)\ge12\)

=> dpcm

Ta có: \(2\left(x^2+y^2+z^2+xy+yz+xz\right)\)

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

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

\(=\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2\)(1)

Mà \(x+y+z=3\Rightarrow\hept{\begin{cases}x+y=3-z\\y+z=3-x\\x+z=3-y\end{cases}}\)

\(\Rightarrow\left(1\right)=\left(3-z\right)^2+\left(3-x\right)^2+\left(3-y\right)^2\)

\(=9-6z+z^2+9-6x+x^2+9-6y+y^2\)

\(=27-6\left(x+y+z\right)+x^2+y^2+z^2\)

\(=9+x^2+y^2+z^2\)

Áp dụng BĐT Cauchy cho 3 số:

\(x^2+y^2+z^2=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}\ge\frac{\left(x+y+z\right)^2}{1+1+1}=\frac{3^2}{3}=3\)

\(\Rightarrow9+x^2+y^2+z^2\ge12\)

hay \(2\left(x^2+y^2+z^2+xy+yz+xz\right)\ge12\)

\(\Leftrightarrow x^2+y^2+z^2+xy+yz+xz\ge6\left(đpcm\right)\)

Đề sai rồi bạn ơi : 

     \(\frac{5^2+6^2}{2}< \frac{\left(5+6\right)^2}{2}\)

Bạn xem lại đề đi.....

Vì x, y cùng dấu nên \(\hept{\begin{cases}\frac{x}{y}>0\\\frac{y}{x}>0\end{cases}}\)

Ta có:

\(\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{y}-2+\frac{y}{x}\right)+2=\left(\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}\right)^2+2\ge2\)

Dấu = xảy ra khi x = y # 0

\(\frac{x}{y}+\frac{y}{x}\ge2\Leftrightarrow\frac{x}{y}+\frac{y}{x}-2\ge0\Leftrightarrow\frac{x^2+y^2-2xy}{xy}\ge0\Leftrightarrow\frac{\left(x-y\right)^2}{xy}\ge0\) luôn đúng!

Thêm phát nữa cho vui

\(Hai.so.duong:\Leftrightarrow\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}\frac{y}{x}}=2\\ \)

\(\frac{1}{1+x^2}+\frac{1}{1+y^2}\ge\frac{2}{1+xy}\)  ( 1 )

\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{1+xy}\right)\ge0\)

\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\frac{y\left(x-y\right)}{\left(1+xy^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\frac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)     ( 2 )

\(\Rightarrow\)Bất đẳng thức ( 2 ) \(\Rightarrow\) Bất đẳng thức ( 1 ) 

( Dấu " = " xảy ra khi x = y ) 

Chúc bạn học tốt !!!

x²+y²z²>=2lxl.lyl.lzl nên VT>=6lxl.lyl.lzl>=6xyz