Find the least value of the expression $x^2+2xy+2y^2+4y+7$ Find the least value of the expression
$x^2+2xy+2y^2+4y+7$
I am not able to solve this equation though i am able to differentiate it. 

Given $x^2+2y^2+2xy+4y+7$$$x^2+2y^2+2xy+4y+7$$$$x^2+y^2+2xy+y^2+4y+7$$$$x^2+y^2+2xy+y^2+4y+4+3$$$$(x+y)^2+(y+2)^2+3$$Minimum of square of any number is $0$. So, minima will be at $x=2,y=-2$
Therefore, $(2+(-2))^2+(-2+2)^2+3=3$
Therefore, the least value of the expression $x^2+2y^2+2xy+4y+7$ is $3$
