Selection Examination A

Since $x$, $y>0$ the given inequality can be written as: $$\begin{array}{rl}(xy+2)(y+2x)\ge 8xy& \iff x{y}^2+2y+2x^{2}y+4x-8xy\ge 0\\ & \iff (x{y}^2-4xy+4x)+(2x^{2}y-4xy+2y)\ge 0\\ & \iff x(y^2-4y+4)+2y(x^2-2x+1)\ge 0\\ & \iff x(y-2{)}^2+2y(x-1{)}^2\ge 0,\text{ which is valid for }x,y>0.\end{array}$$ Equality holds, if and only if: $$x(y-2)=0\text{ and }2y(x-1)=0\iff x=1,y=2,\text{ since }x,y>0.$$

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Alternative solution.

Since $x$, $y>0$ the given inequality can be written as: $$(xy+2)(y+2x)\ge 8xy(1)$$ We apply the well-known inequality of arithmetic – geometric mean $$a+b\ge 2\sqrt{ab},\text{ yea }a,b\ge 0,\text{ (equality when }a=b)\text{ two times, for }a=xy,b=2\text{ and for }a=y,b=2x,\text{ to receive:}$$ $$xy+2\ge 2\sqrt{2xy},\text{(equality for }xy=2\text{)}(2)$$ $$y+2x\ge 2\sqrt{2xy},(\text{equality for }y=2x)(3)$$ From (2) and (3) (multiplication by parts) we get: $(xy+2)(y+2x)\ge 8xy$. Equality holds, if and only if $xy=2$ and $y=2x$, i.e. $x=1$, $y=2$.