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 ?

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.
