If x and y are real numbers, then the minimum value of $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$ is

Expression : $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$

= $$(x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$$

= $$(x + 2y)^2 + \frac{1}{2} (4y^2 - 8y + 8)$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y)^2 - 2 (2y) (2) + (2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y - 2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} (2y - 2)^2 + 2$$

Since, $$x$$ and $$y$$ are real, => Min value of $$(x + 2y)^2 = 0$$

Minimum value of $$(2y - 2)^2 = 0$$

$$\therefore$$ Minimum value of expression = $$0 + 0 + 2 = 2$$