Proving $\frac{x}{\sqrt{y^2+yz+z^2}}+\frac{y}{\sqrt{z^2+zx+x^2}}+\frac{z}{\sqrt{x^2+xy+y^2}} \ge \sqrt{3}$ 
Let  $x,y,z >0$. Prove
  that:$$\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}}
 \ge \sqrt{3}$$

My solution:
By Hölder,
$$\left(\sum\frac{x}{\sqrt{4y^2+yz+4z^2}}\right)^2(\sum x(4y^2+yz+4z^2)) \ge (x+y+z)^3$$
Let us denote $\sum_{sym} x^2y = X$, so we have to prove
$$(x+y+z)^3 \ge 4X+3xyz \iff \sum x^3 +3xyz \ge X$$
which is Schur.
How to prove it in a different way ?
 A: The function $f(x) = \frac 1 {\sqrt x}$ is convex, by Jensen's inequality
$$
\dfrac{x}{\sqrt{y^2+yz+z^2}}+\dfrac{y}{\sqrt{z^2+zx+x^2}}+\dfrac{z}{\sqrt{x^2+xy+y^2}} \geq \\ \frac {(x + y + z)^{3/2}} { \sqrt{x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2)} }
$$
Our inequality becomes
$$
x(y^2 + yz + z^2) + y(z^2 + zx + x^2) + z(x^2 + xy + y^2) \leq \frac 1 3 (x + y + z)^3
$$
Expanding both sides, it is reduced to
$$
xyz\leq \frac {x^3 + y^3 + z^3} {3}
$$
that is true by AM-GM.
A: We need to prove that
$$\sum_{cyc}x\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\sqrt{3\prod_{cyc}(x^2+xy+y^2)}.$$
Now, by C-S 
$$(x^2+xy+y^2)(x^2+xz+z^2)=\left(\left(x+\frac{y}{2}\right)^2+\frac{3}{4}y^2\right)\left(\left(x+\frac{z}{2}\right)^2+\frac{3}{4}z^2\right)\geq$$
$$\geq\left(\left(x+\frac{y}{2}\right)\left(x+\frac{z}{2}\right)+\frac{3}{4}yz\right)^2=\left(x^2+\frac{x(y+z)}{2}+yz\right)^2.$$
Thus, it remains to prove that
$$\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2\geq12\prod_{cyc}(x^2+xy+y^2).$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since $\prod\limits_{cyc}(x^2+xy+y^2)$ is a linear expression of $w^3$, we see that
$$\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2-12\prod_{cyc}(x^2+xy+y^2)=81w^6+A(u,v^2)w^3+B(u,v^2),$$
which says that the inequality
$$\left(\sum_{cyc}(2x^3+x^2y+x^2z+2xyz)\right)^2-12\prod_{cyc}(x^2+xy+y^2)\geq\left(\sum_{cyc}(x^3-x^2y-x^2z+xyz)\right)^2$$
is a linear inequality of $w^3$, which says that it remains to prove the last inequality for an extremal value of $w^3$, which happens in the following cases.


*

*$w^3\rightarrow0^+$.


Let $z\rightarrow0^+$ and $y=1$.
We obtain $$(x-1)^2(3x^4+14x^3+12x^2+12x+4)\geq0,$$
which is obvious;


*$y=z=1$, which gives
$$(x-1)^2(x^3+6x^2+9x+8)x\geq0.$$
Done!

