K gives a great inequality!

$[(x-y)^2+8][(x-y)^2+2] \geq k (x-y)^2$

Applying Cauchy inequality

$[(x-y)^2+8][(x-y)^2+2] \geq (2 \sqrt{2} (x-y) + \sqrt{2} (x-y))^2 =18 (x-y)^2$

So, $k=18$

Let (x-y)^2=t. Thus t>=0 [since it's a perfect square] Now, (x+y)^2=(x-y)^2+4xy=t+4. Thus, our inequality becomes, (t+8)(t+2)>=kt => t+(16/t)+10>=k [since t was positive dividing the inequality with t didn't change sign of inequality] Now the max value of k must be equal to the min value of lhs, so that the inequality holds. Thus if we apply Am-GM inequality in lhs we get value 18.

$For~maximum~it~is~an ~equality. ~~Let~a=x^2+x^{-2}=a.\\ \therefore~\dfrac{(a+4)(a-2)}{a-4}=K.........(A).\\ \therefore~dK/da=\dfrac{(2a^2-6a-8)-a^2-2a+8}{(a-4)^2}=0.\\ So ~a^2=8a.~~a\neq0,~~~\implies~a=8.\\ Substituting~in~ (A)\\ K=\dfrac{12*6} 4 =\huge~~~~~\color{#D61F06}{18}$