Question

x and y are real x^2+2xy-y^2=6. find minimum value of
(x^2+y^2)^2

Answer 1

i wonder, the left looks like a complete square already, or almost

x^2+2xy-y^2=6

x^2+2xy+(y^2-y^2)-y^2=6

(x+y)^2 -2y^2=6

(x+y)^2 = 6+2y^2