Find the minimum of the value $k$ such 
Let $x\geq0$, $y\geq0$, $z\ge 0$ such that
  $$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}\le\sqrt{k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)}$$
  Find the minimum of the real value $k$

I square the side
$$2(x^2+y^2+z^2)+2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}\le k(x^2+y^2+z^2)+(6-k)(xy+yz+xz)$$
$$(k-2)(x^2+y^2+z^2)+(6-k)(xy+yz+xz)\ge 2\sum_{cyc}\sqrt{(x^2+y^2)(y^2+z^2)}$$
Following it seem hard to work
and other hand I think $k_{min}=4\sqrt{2}$,because I let $x=y=1,z=0$,then we have
$$2+\sqrt{2}\le\sqrt{2k+(6-k)}\Longrightarrow k\ge 4\sqrt{2}$$
 A: Yes! For $k=4\sqrt2$ your inequality is true.
Indeed, by C-S
$$\left(\sum_{cyc}\sqrt{x^2+y^2}\right)^2\leq\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}.$$
Thus, it remains to prove that:
$$\sum_{cyc}(x+y+(\sqrt2-1)z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right)$$ or
$$(1+\sqrt2)(x+y+z)\sum_{cyc}\frac{x^2+y^2}{x+y+(\sqrt2-1)z}\leq\sum_{cyc}\left(4\sqrt2x^2+(6-4\sqrt2)xy\right),$$
which is fifth degree homogeneous inequality.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a linear function,
which says that it's enough to prove the last inequality for an extremal value of $w^3$,
which happens in the following cases.


*

*$y=z$.


Since our inequality is homogeneous, we can assume $y=z=1$, which gives
$$(x-1)^2x(x+4\sqrt2)\geq0;$$
2. $w^3=0$.
Let $z=0$ and $y=1$.
We need to prove that
$$(x-1)^2(2(\sqrt2-1)^2x^2+5(3\sqrt2-4)x+2(\sqrt2-1)^2)\geq0,$$
which is obvious.
Done!
