Proving $\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$ The inequality:

$$\left(\frac{a}{a+b+c}\right)^2+\left(\frac{b}{b+c+d}\right)^2+\left(\frac{c}{c+d+a}\right)^2+\left(\frac{d}{d+a+b}\right)^2\ge\frac{4}{9}$$
  Conditions: $a,b,c,d \in \mathbb{R^+}$

I tried using the normal Cauchy-Scharwz, AM-RMS, and all such.. 
I think I can do it using some bash method like expanding the LHS and then maybe to go with homogeneity, normalization and then Muirhead, Jensen or something I dont know since I didn't go that way..
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
 A: Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$, $x+y+z+t=4$, $x+t=2u$ and $y+z=2v$.
Hence, $u+v=2$ and by Rearrangement we obtain: $$\sum\limits_{cyc}\frac{a^2}{(a+b+c)^2}\geq\frac{x^2}{(x+y+z)^2}+\frac{y^2}{(x+y+t)^2}+\frac{z^2}{(x+z+t)^2}+\frac{t^2}{(y+z+t)^2}=$$
$$=\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}+\frac{y^2}{(4-z)^2}+\frac{z^2}{(4-y)^2}$$
and since $$\frac{x^2}{(4-t)^2}+\frac{t^2}{(4-x)^2}\geq\frac{-2(x+t)^2+16(x+t)-16}{(8-x-t)^2} \tag1$$ it's just
$$((x+t)(x^2+t^2)-12(x^2+xt+t^2)+48(x+t)-64)^2\geq0, \tag2$$
it remains to prove that $$\frac{-2u^2+8u-4}{(4-u)^2}+\frac{-2v^2+8v-4}{(4-v)^2}\geq\frac{4}{9},\tag3$$
where $u$ and $v$ are positive numbers such that $u+v=2$ or
$$\sum_{cyc}\left(\frac{-2u^2+8u-4}{(4-u)^2}-\frac{2}{9}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(u-1)(17-5u)}{(4-u)^2}\geq0$$ or
$$\sum\limits_{cyc}\left(\frac{(u-1)(17-5u)}{(4-u)^2}-\frac{4}{3}(u-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(u-1)^2(13-4u)}{(4-u)^2}\geq0.$$
Done!
A: By Holder inequality,we have
$$\left(\sum\dfrac{a^2}{(a+b+c)^2}\right)\left(\sum a(a+b+c)\right)^2\ge \left(\sum_{cyc} a^{\frac{4}{3}}\right)^3$$
therefore
$$\sum\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{(a^{4/3}+b^{4/3}+c^{4/3}+d^{4/3})^3}{[(a+c)^2+(b+d)^2+(a+c)(b+d)]^2}$$
use Holder we have
$$a^{\frac{4}{3}}+c^{\frac{4}{3}}\ge 2\left(\dfrac{a+c}{2}\right)^{\frac{4}{3}}$$
$$b^{\frac{4}{3}}+d^{\frac{4}{3}}\ge 2\left(\dfrac{b+d}{2}\right)^{\frac{4}{3}}$$if setting
$$t^3=\dfrac{(a+c)}{b+d}$$,we have
$$\sum_{cyc}\dfrac{a^2}{(a+b+c)^2}\ge\dfrac{1}{2}\cdot\dfrac{(t^4+1)^3}{(t^6+t^3+1)^2}$$
it remain to show that
$$9(t^4+1)^3\ge 8(t^6+t^2+1)^2,t>0$$
or
$$9\left(t^2+\dfrac{1}{t^2}\right)^3\ge 8\left(t^3+\dfrac{1}{t^3}+1\right)^2$$
let $u=t+\dfrac{1}{t}\ge 2$ the above ineuqality becomes
$$9(u^2-2)^2\ge 8(u^3-3u+1)^2$$
or
$$(u-2)^2(u^4+4u^3+6u^2-8u-20)\ge 0$$
since
$$u^4+4u^4+6u^2-8u-20=u^4+4u^2(u-2)+4u(u-2)+10(u^2-2)\ge 0$$
