Will triangle inequality work ?

Let $a=x+y, b=y+z$ and $c=z+x$ . $a,b,c$ are the sides of a triangle. semiperimeter, $s=\frac{a+b+c}{2}=x+y+z$

$xy+yz+zx= (s-a)(s-b) +(s-b)(s-c) +(s-c)(s-a)=r(r_{a}+r_{b}+r_{c}=r(r+4R)$

In the above i used two results, namely-

i. $(s-a)(s-b)=rr_{c}$ where $r$ is the in-radius and $r_{c}$ is the ex-radius of the ex-circle opposite to vertex C.

ii. $r_{a}+r_{b}+r_{c}= r+4R$ where $R$ is the circum-radius.

$x^2 + xy +y^2$ $=$ $\frac {3(x+y)^{2} +(x-y)^{2}}{4}$ $≥$ $\frac{3}{4}a$

Hence, $r^{2}s^{2}(r+4R){^2}≤\frac{27k}{64}a^{2}b^{2}c^{2}$

or $A(r+4R)≤\frac{3\sqrt{3k}}{8}abc$ [ $A$ denotes the area ]

or $\frac{abc}{4R}≤\frac{3\sqrt{3k}}{8}abc$

or $r+4R≤\frac{3\sqrt{3k}}{8}R$

or $r≤R(\frac{3\sqrt{3k}}{8}-4)$ .

So we require $\frac{3\sqrt{3k}}{8}-4=\frac{1}{2}$ [since $r≤\frac{R}{2}$ ]

Solving this we get $k=3$ .

I think this problem can also be solved with AM-GM.