Using triangle Inequality to solve inequality. Question from a textbook. 
Show that for non negative $x,y,z$ that 
$$ (x+y+z) \sqrt{2}  \leq \sqrt{x^2 + y^2} + \sqrt{ y^2 + z^2} + \sqrt{x^2 +z^2}  $$
and that for $ 0< x \leq y \leq z$,
$$ \sqrt{y^2 +z^2} \leq x\sqrt{2} + \sqrt{(y-x)^2 +(z-x)^2} $$
,which the hint was to utilize the triangle inequality to "an appropriate sum". How does these approaches work? Is there a way to "know" beforehand what these sums are? 

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*$(\sqrt{x^2 + y^2} + \sqrt{ y^2 + z^2} + \sqrt{x^2 +z^2} )^2= 2(x^2 + y^2+ z^2) + \;\;non-negative\;\; terms\geq 2(x^2 + y^2+ z^2)$
which yields the first inequality.


*$(y-x)^2 + (z-x)^2=y^2+z^2+2x^2-2x(y+z)\geq y^2+z^2+2x^2-2\sqrt{2}x(\sqrt{y^2+z^2})=(\sqrt{y^2+z^2}-\sqrt{2}x)^2$
yields the second inequality.
