smallest $k$ s.t. $(x+y)^2\leq k(x^2-xy+y^2)$ I would appreciate if somebody could help me with the following problem
Q: Find $K$? 
$$(x+y)^2\leq k(x^2-xy+y^2)$$.
where $\forall x,y\in \mathbb{R}$

$$(x+y)^2\leq k(x^2-xy+y^2)\iff (k-1)x^2-(2+k)xy+(k-1)y^2\geq0$$
using the inequality
$$xy\leq \frac{1}{2}(x^2+y^2)$$
we have
$$(k-1)x^2-(2+k)xy+(k-1)y^2\geq (k-1)x^2-\frac{2+k}{2}(x^2+y^2)+(k-1)y^2\\
=(\frac{k}{2}-2)(x^2+y^2)\geq 0\iff k\geq 4$$
and if we take $x=y=1$ we find that $4\leq k$.
So  $k= 4$ answers the question.
