Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$ 
Showing $\frac{x+y+z}{\sqrt 2}\le\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}$ $(x,y,z>0)$

Can I apply Jensen here, sum of the squareroots is greater then squareroot of the sum, so I make the RHS smaller then show also then the inequality is valid, i.e. 
$\frac{x+y+z}{\sqrt 2}\le\sqrt{2(x^2+y^2+z^2)}$
squaring both sides we have on the left,
$\left(\frac{x+y+z}{\sqrt 2}\right)^2=\frac{x^2+y^2+z^2+2(xy+yz+xz)}{2}\le\frac{3(x^2+y^2+z^2)}{2}$ 
and on the right we have $2(x^2+y^2+z^2)$, so it is true
If it is OK, can you suggest another proof related more to Cauchy-Schwarz

Recall $\|(x,y,z)\| = \sqrt{x^2+y^2+z^2}$. 
By C-S 
$$x+y+ z= (x,y,z)\cdot (1,1,1) \le \|(x,y,z)\|\|(1,1,1)\|=\|(x,y,z)\|\sqrt{3}.$$ 
But by triangle inequality (follows from C-S): 
$$ \|(x,y,z)\| = \frac 12  \|(x,y,0) +(0,y,z)+ (x,0,z)\|\le \frac 12 \left(\|(x,y,0)\|+\|(0,y,z)\|+\|(x,0,z)\|\right).$$ 
Summarizing: 
$$(*)\quad x+y+z \le \frac{\sqrt{3}}{2}\left ( \|(x,y,0)\|+\|(0,y,z)\|+\|(x,0,z)\|\right).$$ 
But $\frac{\sqrt{3}}{2}<1<\sqrt{2}$, and we're all set. 
