How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$ let $x,y,z$ are postive numbers,show that
$$\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}+\dfrac{y+z}{\sqrt{y^2+yz+z^2+zx}}+\dfrac{z+x}{\sqrt{z^2+zx+x^2+xy}}\ge 2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$
My try: Without loss of generality，we assume that $$x+y+z=1$$
and use Holder inequality,we have
$$\left(\sum_{cyc}\dfrac{x+y}{\sqrt{x^2+xy+y^2+yz}}\right)^2\left(\sum_{cyc}((x+y)(x^2+xy+y^2+yz)\right)\ge\left(\sum_{cyc}(x+y)\right)^3$$
then we only prove this
$$\dfrac{8}{\sum_{cyc}(x+y)(x^2+xy+y^2+yz)}\ge  2+\sqrt{\dfrac{xy+yz+xz}{x^2+y^2+z^2}}$$
then I can't
Thank you
 A: We can use here the Ji Chen's lemma: 
https://artofproblemsolving.com/community/c6h194103
for $p=\frac{1}{2}.$
For which it's enough to prove that:
1.$$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq2+\frac{xy+xz+yz}{x^2+y^2+z^2},$$
2.$$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq1+\frac{2(xy+xz+yz)}{x^2+y^2+z^2},$$ 
3. $$\prod_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq\frac{xy+xz+yz}{x^2+y^2+z^2}.$$

A proof of the first inequality.

By C-S $$\sum_{cyc}\frac{(x+y)^2}{x^2+xy+y^2+yz}\geq\frac{\left(\sum\limits_{cyc}(x+y)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2+yz)}=\frac{2\sum\limits_{cyc}(x^2+2xy)}{\sum\limits_{cyc}(x^2+xy)}.$$
Now, let $x^2+y^2+z^2=k(xy+xz+yz).$
Thus, $k\geq1$ and it's enough to prove that $$\frac{2(k+2)}{k+1}\geq2+\frac{1}{k}$$ or $$k\geq1,$$ which is true;

A proof of the second inequality.

By C-S and Rearrangement $$\sum_{cyc}\frac{(x+y)^2(y+z)^2}{(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}\geq\frac{\left(\sum\limits_{cyc}(x+y)(y+z)\right)^2}{\sum\limits_{cyc}(x^2+xy+y^2+yz)(y^2+yz+z^2+zx)}=$$
$$=\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+2x^3y+3x^3z+4x^2y^2+6x^2yz)}=$$
$$=\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)-xyz\sum\limits_{cyc}\left(\frac{x^2}{z}-x\right)}\geq$$
$$\geq\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)}.$$
Id est, it's enough to prove that
$$\frac{\left(\sum\limits_{cyc}(x^2+3xy)\right)^2}{\sum\limits_{cyc}(x^4+3x^3y+3x^3z+4x^2y^2+5x^2yz)}\geq\frac{(x+y+z)^2}{x^2+y^2+z^2}.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, it's obvious that the last inequality is a linear inequality of $w^3$, which says that it's enough to prove the last inequality for an extreme value of $w^3$, which happens in the following cases.
a) $w^3\rightarrow0^+$.
Let $z\rightarrow0^+$ and $y=1$.
We need to prove that $$\frac{(x^2+3x+1)^2}{x^4+1+3x^3+3x+4x^2}\geq\frac{(x+1)^2}{x^2+1}$$ or
$$x(x^4+x^3-2x^2+x+1)\geq0,$$ which is obvious;
b) Two variables are equal. 
Let $y=z=1$.
Here we obtain:
$$(x-1)^2(2x^3+11x^2+12x+2)\geq0.$$

A proof of the third inequality.

We need to prove that
$$(x^2+y^2+z^2)\prod_{cyc}(x+y)^2\geq(xy+xz+yz)\prod_{cyc}(x^2+xy+y^2+yz),$$
which is obviously true after full expanding:
$$\sum_{cyc}(x^6y^2+x^5y^3+x^6yz+2x^5y^2z-x^4z^3y-x^4y^2z^2-3x^3y^3z^2)\geq0.$$
Done!
