<s>[INST] 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
 [/INST] Hint: Note that $ \sqrt{2(a^2+b^2)} \geq a+b, ~\forall a,b\in \mathbb{R}$ and use it to derive the desired inequality. 
</s>